Best lavalier, lapel mic, the ultimate guide – cheapest to the best money can buy right now.

Lavalier, lapel mics:

Sennheiser MKE-40 

Sennheiser MKE 40-EW

Sennheiser MKE 40: A Long-Standing Lavalier Microphone

The Sennheiser MKE 40 is a professional lavalier microphone renowned for its durability and high-quality audio. While there’s only one core design, the MKE 40 can be configured for wired or wireless use.

Production History:

  • Introduced in 1981, the MKE 40 was a pioneering lavalier microphone with a cardioid polar pattern, offering superior sound directionality compared to earlier omnidirectional lavalier mics. (
  • The core design has likely benefitted from improvements over the years. Technological advancements in miniaturization, audio processing, and materials science would likely prompt Sennheiser to incorporate them gradually into the MKE 40.
  • Manufacturers often make incremental improvements without announcing new variations. These could encompass better internal components, minor design tweaks, or manufacturing efficiencies.
  • One confirmed variation is the MKE 40-1053, distinguished by its LEMO connector, suggesting variations might have existed based on connector types. (


  • The standard MKE 40 likely uses a different connector compared to the MKE 40-1053 variation.
  • The MKE 40-EW designation likely refers to the standard MKE 40 bundled with a compatible Sennheiser wireless transmitter/receiver for a complete wireless setup, not a separate microphone variation.

Overall, the Sennheiser MKE 40’s longevity speaks to its quality and adaptability. While specific details about hardware upgrades might be challenging to find, the timeline suggests it has likely been refined to incorporate advancements in microphone technology.

Sennheiser MKE 40: Technical Details

Here’s a breakdown of the Sennheiser MKE 40’s technical details and a comparison to similar lavalier microphones:

MKE 40:

  • Impedance: While specific impedance specifications are difficult to find for the MKE 40, lavalier microphones typically range from 100 ohms to 1000 ohms. The MKE 40 likely falls within this range, but an official source is needed for confirmation.
  • Connectors:
    • Standard MKE 40: Likely uses a connector different from the LEMO connector of the MKE 40-1053 variation.
    • MKE 40-1053: LEMO connector.
  • Quality Comparison:

The MKE 40 is known for: * Cardioid polar pattern: Offers good directionality, reducing unwanted ambient noise. * Durability: Built to withstand professional use. * Discreet size: Clips easily onto clothing.

Here’s how it compares to similar lavalier microphones:

  • Rode Lavalier GO: More affordable option, omnidirectional pattern (picks up more ambient noise).
  • Shure MVL: Similar price range, omnidirectional pattern, known for clarity.
  • Sennheiser MKE 600: Higher-end option, supercardioid pattern (even tighter directionality), improved sensitivity.

Choosing a Lavalier Microphone:

The best microphone depends on your needs. Consider:

  • Budget: The MKE 40 falls in a mid-range price bracket.
  • Polar pattern: Cardioid offers a good balance of directionality and sound quality.
  • Durability: The MKE 40 is known for being robust.
  • Connector compatibility: Ensure compatibility with your recording equipment.

Connector Variations:

  • Current Variants:
    • MKE 40-EW: This widely available version features a standard 3.5mm jack connector, making it compatible with a vast range of recording equipment.
    • MKE 40-4: This variant utilizes a 3-pin LEMO connector, commonly used in professional audio applications. It offers a secure and reliable connection for demanding environments.
  • Archived Variant:
    • MKE 40-1053: This older version comes equipped with a single-pin LEMO connector, which is no longer in mainstream use. While Sennheiser doesn’t offer an official converter/adapter, you might be able to source a single-pin LEMO to 3.5mm jack cable from third-party vendors.

History (Limited Information Available):

While details about specific upgrades are scarce, Sennheiser support acknowledges the MKE 40’s long production history, suggesting it likely benefitted from advancements in microphone technology over time.


Additional Notes:

  • The MKE 40’s design prioritizes durability and high-quality audio capture, making it a popular choice for professional applications.
  • The cardioid polar pattern provides good directionality, reducing unwanted background noise.
  • When choosing an MKE 40 variant, consider your recording equipment’s compatibility with the connector type (3.5mm jack vs. 3-pin LEMO).

Further Exploration:

  • Searching for vintage microphone resources or contacting Sennheiser directly might yield more insights into the MKE 40’s historical development.
  • Online reviews and comparisons with similar lavalier microphones from Rode, Shure, and Sennheiser’s higher-end MKE 600 can offer a comprehensive picture of the MKE 40’s strengths and limitations.

Connectors (Depending on Version):

  • 3-pin LEMO connector (MKE 40-4)
  • 3.5mm jack connector (MKE 40-EW)


  • MZW 42 Windscreen: Available in grey and black.
  • Magnetic mount set: This might be referred to as MZM 2 / MZM 10 in older documents.
  • Clip: Color not confirmed from the manual, but the image shows a black clip.

Sennheiser Clamp MKE40


Reserve onderdelen


Capsule, electret, cardioid

Capsule, electret, cardioid


Cable with Lemo plug

Cable with Lemo plug


Clip for MKE 10/MKE 40

Clip for MKE 10/MKE 40








Parts Not Confirmed by Manual:

  • Single-pin LEMO connector (MKE 40-1053 – archived version)
  • Converter/adapter for single-pin LEMO connector (not offered by Sennheiser)

Manual MKE_40_Instruction_Manual_10_2021.pdf


Sennheiser MZW 42 Windshield black for MKE-40 

Sennheiser MKE 40-1053 Ansteckmikrofon mit LEMO-Stecker

Sennheiser MKE 40 1053 dasspeld microfoon

Sennheiser MKE 40-EW dasspeldmicrofoon

Sennheiser MKE 40 1053 dasspeld microfoon

Sennheiser MKE 40-1053 lemo single pin

Lemo 1 pin koax connector for bodypacks BF 1083-U/BF 1083-V

The Sennheiser MZW 42 is a foam windscreen designed specifically for use with Sennheiser’s ME 42, ME 62, ME 64, and ME 66 microphone capsules. Here are some useful aspects of the Sennheiser MZW 42:

  1. Wind Noise Reduction: The primary purpose of the MZW 42 is to reduce wind noise when recording outdoors or in environments with air movement. The foam material helps to diffuse wind gusts, preventing them from causing distortion in the audio.
  2. Protection from Moisture and Dust: In addition to reducing wind noise, the MZW 42 also provides a degree of protection against moisture and dust. This can be particularly useful when recording in outdoor environments where weather conditions may be unpredictable.
  3. Compatibility: The MZW 42 is specifically designed to fit Sennheiser’s ME 42, ME 62, ME 64, and ME 66 microphone capsules. It’s essential to ensure compatibility with these microphone models before purchasing.
  4. Durable Construction: The foam material used in the MZW 42 is typically durable and long-lasting, providing reliable protection for your microphone capsule over time.
  5. Easy to Install: The windscreen is designed for quick and easy installation onto the microphone capsule. It simply slides over the microphone, providing instant wind protection without the need for complex setup procedures.
  6. Compact and Portable: The MZW 42 is lightweight and compact, making it easy to transport and use in various recording situations. It’s a practical accessory for field recording and outdoor videography where portability is essential.

Overall, the Sennheiser MZW 42 foam windscreen is a valuable accessory for Sennheiser microphone users, offering effective wind noise reduction and protection in outdoor recording environments.

Sennheiser MKE 40-EW Clip on microfoon

SENNHEISER MKE 40-4 MICROPHONE Clip-on, cardioid, 3-pin Lemo connector, black

Parts for connection

Alla spare parts on sennheiser website

Lemo single pole connector


HB Lemo 3p Contra > 20cm > Mini jack Schoef

00B 3pin connector FVB.00.303.NLAE24 for Sennheiser SK 50/250/2000/5012/6000

LEMO FFA00250CTA HF-kabelstekker, NIM-CAMAC

LD Systems U500 GC


pro snake 20063

lemo fvb.00

Sennheiser MZW 42 Wind Protection for MKE 40 EW Lavalier Microphone

Sennheiser CL2 – Line Cable

SENNHEISER MKE 2 (BLACK 3-PIN) Clip-on microphone, omnidirectional, 3-pin SE connector

Oversized lapel mic used on a BBC FIFA story…A Sennheiser MKE40-4

MKE 40-4

Sennheiser MZA 900P

preamp / phantom adapter (XLR – 3,5mm jack)

SD Systems C-SE Sennheiser

Voor het gebruik van SD Systems microfoons met Sennheiser EW draadloze systemen

Sennheiser MZA 900P

senheisse MKE 40 phantom power

LR to Microphone wiring

connecting balanced wire to trs plug



MKE 40 sounded best and rejected more of the extraneous background noise.

2 – AT 831

3 – Rode

4 – ME2

Sennheiser MKE 2 (Black EW)

Sennheiser MKE 1-4


Sanken COS-11D

Technical details of the Sanken COS-11D:

  1. Impedance (Ohms): The COS-11D typically has an impedance of around 300 ohms. This impedance level is standard for lavalier microphones and ensures compatibility with various audio equipment.
  2. Connectors: The COS-11D comes with different connector options to suit different wireless transmitter systems. Common connector types include TA5F, 3.5mm locking connector, and various wired connectors such as XLR. These connectors ensure compatibility with a wide range of audio recording devices.
  3. Audio Quality: The COS-11D is known for its exceptional audio quality, characterized by clear and natural sound reproduction. Its omnidirectional pickup pattern captures sound from all directions evenly, ensuring consistent audio quality regardless of the speaker’s position.
  4. Build Quality: Sanken is renowned for its high build quality and durability, and the COS-11D is no exception. It features a robust construction designed to withstand the rigors of professional use, including moisture resistance for outdoor recording environments.
  5. Frequency Response: The COS-11D typically has a frequency response range of around 50Hz to 20kHz. This wide frequency response captures a broad range of audio frequencies, resulting in accurate and detailed sound reproduction.
  6. Sensitivity: The sensitivity of the COS-11D varies depending on the specific model and connector type but generally falls within the range of -40 dB (0 dB = 1V/Pa). This sensitivity level ensures adequate signal levels for clear audio capture without the need for excessive gain.

Overall, the Sanken COS-11D lavalier microphone stands out for its excellent audio quality, durable construction, and compatibility with a wide range of recording equipment, making it a preferred choice for professional audio applications.

DPA 4060-OC-C-B00

DPA 4060-OC-C-B00: A Miniature Lavalier Microphone

The DPA 4060-OC-C-B00 is a miniature lavalier microphone designed for professional applications such as broadcast, performance, and theatrical use. Here’s a breakdown of its key features based on information readily available online:

Form Factor: Lavalier – small and lightweight for discreet clipping onto clothing.

Polar Pattern: Omnidirectional – picks up sound equally from all directions.

Frequency Range: 20 Hz to 20 kHz – captures a wide range of audible frequencies.

Maximum SPL: * 129 dB SPL (Plug-In Power, 1% THD) * 134 dB SPL (Plug-In Power, 10% THD) – Handles high sound pressure levels without distortion.

Power Source: Plug-In Power – requires 48V phantom power from a compatible audio system.

Color: Black

Connector: Microdot Coaxial Male – This is a small, low-profile connector commonly used with lavalier microphones.

Additional Notes:

  • DPA’s CORE technology is likely used in this model, offering a 6 dB increase in dynamic range and a 6 dB reduction in noise compared to the previous 4060 Legacy model (information not always explicitly mentioned for all variations).
  • DPA offers various microphone capsule and connector options within the 4060 series. The “OC-C-B00” designation likely refers to a specific combination of these options.
  • Without the full model number details or consulting the official DPA website/manual, it’s difficult to pinpoint the exact capsule type or presence of features like a presence boost.

Similar microphones to consider:

  • Sennheiser MKE 40: Offers a cardioid polar pattern for more directional sound pickup.
  • Shure MVL: Known for clarity and omnidirectional pattern.
  • Rode Lavalier GO: More affordable option, omnidirectional pattern.

For comprehensive information:

  • Always refer to the official DPA website or product manual for the most accurate specifications of the DPA 4060-OC-C-B00, including details about specific capsule types, presence boost features (if applicable), and available accessories.

Sennheiser Headphone Pre Amps

1 pin lemo naar jack

TQG (4-pole) to 6.3 mm right angle mono jack Length: 60 cm Color: black

Shure WA 304


Boya BY-M1 Lavalier Microfoon

The BOYA BY-M1 Lavalier Microphone is a popular choice for recording audio in various situations such as interviews, vlogging, presentations, and more. Here are some useful features and aspects of this microphone:

  1. Omni-directional Pickup Pattern: The microphone has an omnidirectional pickup pattern, which means it captures sound from all directions. This feature is particularly useful for capturing sound in environments where the source may move around or where there are multiple speakers.
  2. Compatibility: The BOYA BY-M1 is compatible with a wide range of devices including smartphones, DSLRs, camcorders, audio recorders, and PCs. This versatility makes it a practical choice for different recording setups.
  3. Long Cable: The microphone comes with a lengthy cable, allowing for more flexibility in positioning the microphone. This is especially handy if you need to move around during recording without worrying about being restricted by a short cable.
  4. Clip-on Design: Its clip-on design makes it easy to attach the microphone discreetly to clothing, making it ideal for interviews and situations where visibility of the microphone needs to be minimized.
  5. Value for Money: The BOYA BY-M1 is often praised for its affordability compared to other lavalier microphones on the market. Despite its lower price point, it still offers decent sound quality and reliability for most recording needs.
  6. Accessories: The microphone usually comes with additional accessories such as a foam windscreen and a 1/4″ adapter, which can be useful for reducing wind noise and connecting to different types of audio equipment.

Overall, the BOYA BY-M1 Lavalier Microphone is a budget-friendly yet reliable option for anyone

Connecting stereo jack mic to headset for android phone:
Three pin 3.5mm jack is called TRS
Four pin 3.5mm jack including mic pin is called TRRS

3.5mm TRS to TRRS RODE SC4 Adapter Cable Compatible with RODE Wireless GO II Wireless Microphone Connection to Mobile Phone Adapter

Rode SC4

Rode SC4

The Rode SC4 is a high-quality shielded adapter cable designed to allow microphones with a 3.5mm TRS output to connect to TRRS smartphones and tablets. This is particularly useful for content creators, journalists, and anyone else who needs to record audio directly into a smartphone or tablet for video production, interviews, podcasts, and other applications.

The SC4 essentially converts the microphone’s output from a TRS (Tip-Ring-Sleeve) connection to a TRRS (Tip-Ring-Ring-Sleeve) connection, which is the standard for most smartphones and tablets. By using the SC4, you can ensure compatibility between your microphone and your mobile device, allowing you to capture high-quality audio without the need for additional adapters or converters.

pro snake MC 4

Replacement SC3 microphone cable for Red 3.5mm TRRS to TRS adapter for cameras, videorecorders compatible with Sony, Nikon, Canon.

Trs to trrs

Pro snake MC 4

Jack verloopstekker

Search link for lapel mic:

Connector jack TRS shielded to connect to lavalier mic:

Neutrik NTP3 RC

Zoom H4n Pro Black

Audio recorders for lavlaier lapel mic

Zoom H4n Pro Black

Multi connectors

pro snake 83840-5 Multicore

The Pro Snake 83840-5 Multicore is a type of audio cable commonly used in professional audio setups, particularly in live sound and recording environments. “Multicore” refers to a cable that contains multiple individual cables within a single outer jacket. These individual cables are often color-coded and individually shielded to minimize crosstalk and interference between channels.

The “83840-5” designation likely refers to the specific model number or product code for this particular multicore cable manufactured by Pro Snake.

Multicore cables are frequently used to connect audio equipment over distances, such as from a mixing console to stage monitors, amplifiers, or recording devices. They offer the advantage of reducing cable clutter and simplifying setup by consolidating multiple connections into a single cable bundle.

The Pro Snake brand is known for manufacturing a range of audio cables and accessories, with a focus on quality and reliability for professional audio applications.

Mini jack shielded

Best microphones videos and info

Best Lapel Microphones

Trs pins L R S

Connectinf balanced lavalier mice to jack pin 3.5 mm

Vervanging 3.5Mm Trs Naar 3.5Mm Trrs Adapter Kabel Voor Rode SC7 SC2 Door Video Mic Draadloze Go Video micro-Type Microfoons

Sennheiser dissamble:

No sound from bluetooth speaker, not starting? Easy solution.

If you are experiencing no sound from your Bluetooth speaker, there are a few things you can try to troubleshoot the issue:

  1. Check the volume: Make sure that the volume on the speaker and your device (such as a phone or computer) are turned up.
  2. Check the connection: Ensure that the speaker and your device are properly paired and connected.
  3. Restart the device and speaker: Turn off the speaker, and then turn it back on. Then restart your device and try to reconnect it to the speaker.
  4. Check the battery: If you are using a portable Bluetooth speaker, check to make sure the battery is charged.
  5. Check if the speaker is on pairing mode: Some speakers have a button to activate pairing mode. Make sure the speaker is not in pairing mode.
  6. Check your device’s Bluetooth settings: Go to the Bluetooth settings on your device and check if the speaker is listed as connected. If not, try to reconnect it.
  7. Check other apps: Close other apps that might be using the audio on your device and try again.
  8. Update the speaker’s firmware: Some manufacturers release firmware updates to fix sound issues. Check the manufacturer’s website for updates.

If the above steps don’t work, it may be a hardware issue, and you may need to contact the manufacturer for further assistance.

For example for JBL Charge, Boombox or Flip contact JBL

What are the most famouse bluetooth speakers? Click here to learn more.

Troubleshooting Electronic Circuits

Troubleshooting Electronic Circuits

Quick Notes: Replacing Electrolytic Capacitors and Soldering

For now, here’s a short summary on troubleshooting techniques for the experienced hobbyist concerning (aluminum) electrolytic capacitors.

If you are restoring older electronic devices such as radios, tape recorders, ste- reos, power supplies, signal generators, etc., you should most likely replace all the electrolytic capacitors with the same (or sometimes larger) capacitance and voltage rating. For example, if there is a 100 μf, 10-volt electrolytic capacitor that has lost its capacitance or became leaky, you can replace it with a new 100 μf, 16-volt version. Of course, make sure that the replacement capacitor is installed correctly in terms of

polarity. Also, if the device runs off the power outlet (e.g., 110 volts AC or 220 volts AC), then make sure you completely disconnect its 110v/220v power cord from the wall (or mains) outlet before repairing the device.

Even in many newer devices, you can spot a bad electrolytic capacitor by notic- ing if the case is bulging. For example, normally the top of an aluminum electrolytic capacitor is flat, but a bad one might be curved up like having a mound added on top of the capacitor. Also, if there are chemical residues near the leads (e.g., white or blue powder, gel, or liquid), these are telltale signs of a bad electrolytic capacitor. See Figure 1-1 that shows a residue leak.

FIGURE 1-1 A bad capacitor on the left side (A), and a good capacitor on the right side (B).

Should you decide to repair or restore older electronic devices from the 1950s to 1980s, keep in mind that aluminum capacitors have about a 15-year life. However, if they are in a warmer environment such as in a TV camera, electrolytic capacitors can fail within 5 years.

Soldering Problems (Cold Solder Connections)

If you are starting out in electronics, then you can build circuits with solderless breadboards, solder your own circuits, or purchase preassembled soldered boards. For troubleshooting, sometimes you will be required to solder when working with new circuits, replacing components, or repairing/modifying printed circuit assem- blies. When soldering connections, we must avoid making cold solder joints where the solder may have been “tacked” or “dabbed” onto the circuit, or the solder had not melted sufficiently. See Figure 1-2.

In Figure 1-2, we see that all connections can be improved by just reheating and adding a little more solder until the connections create a smooth mound of solder. If your connections have a “spiky” look similar to number 3 in Figure 1-2, then you may only have a temporary connection that can easily be disconnected with a slight pull on the wire. Generally, spend an extra few seconds on the solder joint to allow the solder to flow all around the connection. See Figure 1-3.

FIGURE 1-2  Bad solder connection examples.

FIGURE 1-3 A good connection where the solder has been heated sufficiently to flow properly.


Throughout this book we will include basic electronics circuit theory such as Ohm’s Law applied to the associated circuits that are discussed.

Finally, do not forget to read Appendix A, which covers choosing test equipment (e.g., power supplies, oscilloscopes, and generators) and shows some of their limita- tions.

And now let’s proceed to Chapter 2.

               CHAPTER 2

Basic Breadboards

This chapter will examine various breadboards for constructing circuits. We will cover solderless breadboards, copper clad bare printed circuit boards, and perfo- rated or vector boards.

Let’s first look at solderless breadboards, which can vary in size and quality.

Solderless Breadboards

In Figures 2-1 and 2-2, we see two different types of solderless breadboards. Note that each of them has tabs (e.g., Tab1, Tab2, and Tab3) that allow multiple boards of the same type/size to be expanded.

Let’s start with Figure 2-1’s solderless breadboard. In each column, there is a column number (e.g., 1 to 30) and five rows (a to f and g to l).

Each column (e.g., column 1 to 30) is connected from rows a to f in the upper portion of the breadboard. And on the lower portion each column (e.g., column 1 to 30) from rows g to l is connected.

The columns 1 to 30 from the upper portions with rows a to f are independent and not connected to any of the columns in the lower portion, such as columns 1 to 30 and row g to l.

Stated in other words, each column is independent and not connected to any other column, even if they are on the upper or lower portion of the breadboard. For example, column 1 is insulated and not connected to any other column such as col- umn 2.

For the example in Figure 2-2, each column from 0 to 60 on the top rows, and on the lower rows the five locations in rows A to E and F to J are connected together. For example, the hole in column 1 row A is only connected to column 1 rows B, C, D, and E. The same connections apply to the other columns, for example, column 3 row F is only connected to column 3 rows G, H, I, and J.


FIGURE 2-1  A small 30-column solderless breadboard with wire jumper connectors.

FIGURE 2-2 A larger-sized solderless jumper board with 60 columns. The power buses are denoted by the red and blue lines running across the solderless breadboard.

There are two power buses denoted by red and blue lines toward the edge of the board. We will discuss power buses in more detail later; some are broken into “sectors” within the board, and others are not.


Figure 2-3 shows a high-quality solderless breadboard where the connectors inside the wells or holes do not cause wire or leads from electronic parts to be jammed.

As we can see in Figure 2-4, the holes or wells labeled “Bad” are closed off, leav- ing less area to insert wire leads from components. The metal connector clips inside the holes should be expanded outward to allow a larger hole or well size for easier lead or wire insertion.

FIGURE 2-3 A high-quality solderless breadboard where the wells or holes are clear and allow for easy insertion of wire leads from electronics components such as resistors, capacitors, etc.

FIGURE 2-4 A lower-quality solderless breadboard with internal connectors or “blades” narrowing the pathways for inserting wires or electronic components.

So be on the lookout for these and avoid them because plugging in resistors, capacitors, and wires will be difficult. One possible workaround on this is to order parts (e.g., resistors and capacitors) with thinner leads. However, you may find that the standard solderless jumper wires may have difficulty plugging in with these par- tially blocked or narrowed wells.

You can order boards that are made by well-known manufacturers such as Bud Industries, BusBoard, and Twin Industries, or electronics vendors such as Jameco, Mouser, Digi-Key, and Adafruit. Should you order the solderless breadboards from Amazon, look for the user ratings before buying. If you order via eBay, you may find some that are not so good.

Power Buses on Solderless Breadboards . . . Look for Breaks in the Power Bus Lines

Some solderless breadboards will partition one or more power buses to allow differ- ent voltages to be applied. For example, a digital and analog system may require +5 volts and +12 volts, respectively. The ground or minus connection may be common to both power supplies. See Figure 2-5.

FIGURE 2-5 A solderless breadboard with sectioned or separated power buses as denoted by a break in the red/gray lines.

The solderless board in Figure 2-5 shows that the power bus used for ground or common power supply minus lead is continuous as denoted by the black lines from columns 1 through 30. Columns 14 and 17 are internally connected and are marked in Figure 2-5 by “—–C  ” in black font overlays.

However, the power bus with the red/gray lines from columns 1 to 14 is not con- nected to columns 17 to 30 and there are “no connects” denoted by “—NC  ” in

red/gray font overlays.

In this example, it is possible to have up to four different power supply voltages with a common connection or ground connection.

To be sure, you should use a continuity tester or ohm meter to confirm no con- nections between columns that are marked to be separated, which will show “infi- nite ohms” for the two columns 14 and 17 in red/gray indicated by “—NC—”. The two columns 14 and 17 in black with “—-C—” should measure continuity or close to 0 ohms.

Figure 2-5 is not indicative of all solderless breadboards in terms of separate buses. For example, Figure 2-6 shows a double-size board with continuous buses throughout.

FIGURE 2-6  A solderless breadboard without separations in the “+” and “–” bus lines

One reason for having separate buses is so that a larger board can have multiple sectors with different types of circuits requiring different power supply voltages. In general, you should always test for continuity on all buses around the middle of the board.

Figure 2-7 shows a board with four “independent” sectors.

If you want to connect the power buses together, you can use insulated or un- insulated wires as shown in Figure 2-8.

See Figure 2-9 for a close-up of the bare wires connecting a couple of the power buses.

FIGURE 2-7 A board with four sectors and independent power bus lines as denoted by the breaks in the six pairs of lines at columns 32 and 33.

FIGURE 2-8  Bare (un-insulated) wires connecting the power buses.

FIGURE 2-9 Bare wires connecting the two power buses.

And Now Some Words of Caution

If the power bus lines are mismarked or misinterpreted in any way, damage to your circuits or power supply may occur. For example, if you think that the power buses are connected in the middle and apply power, you may be only powering half the board.

On the other hand, if the board is marked with a break in the power bus lines, but in fact they are really connected, then when you apply different voltages on the allegedly different power buses, you may short-circuit one supply to another, or send too high of a voltage into circuits that cannot tolerate the high voltage. For example, if you have +5-volt and +12-volt supplies, then the +12-volt supply could send 12 volts throughout the board and into digital circuits that are normally running at 5 volts. With this example, digital chips will be “fried” or damaged, and worse yet it can cause smoke from the chips, or even injury. See Figure 2-10.

The user should confirm continuity on the top board at columns 29 and 31 for the “–” and “+” buses, which were confirmed to be connected for all buses. The top board is incorrectly marked and should have had solid bus lines drawn like in the middle and bottom boards.

This is why before you apply power to your solderless breadboard, you should check the power buses for continuity in the middle of the board.

FIGURE 2-10 All three boards have continuous bus connections, including the top board that has breaks in the “–” and “+”bus lines that would have indicated separated “–” and “+” bus lines (red and blue).

Other Breadboards

For analog circuits up to about 10 MHz, including shortwave radios, I have been able to produce satisfactory results with solderless breadboards. For higher frequency circuits, use a copper clad board. See Figure 2-11.

For high-frequency circuits up to 500 MHz, copper clad breadboards work fine. Figure 2-11 shows on the left side a circuit made with through-hole parts. For very high frequencies, sometimes you can use smaller parts, including surface mount resistors, capacitors, and semiconductors. Some of these components will be covered in the subsequent chapters.

FIGURE 2-11 Left side is a “dead bug” style copper clad construction and right side shows a blank copper clad printed circuit board.

If you are building logic circuits, then solderless breadboards can work for low- speed logic gates such as CD4000 series or 74Cxx CMOS gates. If you use faster gates such as 74HCxx gates, you will have to keep track of pulse glitches from the chips propagating through the ground and/or power supply bus. One way to filter out the glitches is to slow down the rise/fall times via an RC filter at the output of each gate. But these RC networks can add up to too many.

The best alternative then is to build on a blank printed circuit board, a vector board with ground plane, or to lay out a PC board. For the hobbyist, usually the vec- tor board (perforated board) approach is a good balance. See Figure 2-12.

FIGURE 2-12 A perforated board with a ground plane.

Perforated boards can be made with fiberglass such as the one shown in Figure 2-12, or you can buy more inexpensive phenolic types.

The hole spacing in vector or perforated boards is usually 0.100 inch or 100 mils. This is a standard spacing for many through-hole integrated circuits, connectors, LEDs, transistors, and jumper terminals.

               CHAPTER 3

Power Sources:

Batteries and Battery Holders, Safety Issues,

and Voltmeters

We first cover different types of batteries and their holders. In particular, some hold- ers have similar connectors and can cause the wrong voltage to be supplied to your circuits. Also, we will take a look at various digital voltmeters that will be one of your main troubleshooting tools.


Many circuits today can be powered by about 3 volts to 9 volts. These voltages can be supplied by batteries connected in series. Although there are different types of 1.5- volt battery sizes, such as AAA, AA, C, and D, the most common ones to use are AA. There are the shorter-life carbon zinc or manganese batteries, and longer-life alka- line cells. All of these are “primary” cells, which must be discarded after they run out. That is, they are not rechargeable!

There are “standard” rechargeable batteries, such as nickel cadmium (NiCd) and nickel metal hydride (NiMh). However, these batteries will lose their charges within a few months.

A better investment is to buy “low self-discharge” nickel metal hydride (NiMh) batteries. Low self-discharge rechargeable batteries will retain about 80 percent capacity one year after a full charge. The typical capacity of these cells is about 2000 milliamp-hour for the AA, 4000 milliamp-hour for the C, and 8000 milliamp-hour for the D size.

Battery capacity in milliamp-hour gives us a good estimate of battery life based on current drain. For example, suppose you are powering an LED flashlight with an


AA NiMh rated at 2000 milliamp-hour, and the LED flashlight drains 200 milli- amps. We can then calculate the flashlight’s duration to give off light as:

time or duration = (battery capacity)/(current drain)

In this example, battery capacity for an AA cell is 2000 milliamp-hour, and cur- rent drain is 200 milliamp.

time or duration = (2000 milliamp-hour)/(200 milliamp) = 10 hours time or duration = 10 hours

Rechargeable NiMh 9-volt batteries also come in the low self-discharge variety.

Usually, the other key words to look out for in low self-discharge rechargeable batteries are

•     “Ready to use”

•     “Precharged” or “Pre-Charged” or “Factory Charged”

•     “Charge Retention, up to 12 months” or “keeps charged up to 12 months”

•     “Long shelf life”

Here are a few manufacturers of rechargeable long–shelf life batteries:

•     Eneloop by Panasonic or Sanyo

•     energyOn

•     Eveready

•     Rayovac

Note that NiMh batteries give about 1.25 volts per cell, but fortunately, most electronic circuits will operate in the 2.5-volt (2 cells in series) to 7.2-volt (6 cells in series) range.

See Figure 3-1, which shows various double layered AA cell battery holders.

FIGURE 3-1 AA battery holders for 4, 6, 8, and 10 cells.

Note the battery holders’ positive (+) and negative (–) polarity markings. The smaller connector on the left side of each battery holder is the positive terminal, while the larger diameter connector toward the right side is the negative terminal.

See Table 3-1 for a voltage summary on these holders.

TABLE 3-1 Nominal Output Voltages Based on Number of AA Cells

Number of CellsVoltage Rechargeable NiMhVoltage Alkaline or Carbon
45.0 volts6.0 volts
67.5 volts9.0 volts
810.0 volts12.0 volts
1012.5 volts15.0 volts

Some battery holders come in single layer construction. See Figure 3-2.

FIGURE 3-2 Flat battery holders with leads (red wire is positive and black wire is negative) and with connectors.

Battery holders with leads are preferred sometimes so that there is no chance of accidentally supplying the wrong voltage to a specific circuit. The reason is because if you set aside a circuit for a while, you may forget what the circuit voltage is. If it has a “9-volt” connector, then the first assumption is to connect 9 volts to the circuit, which may burn it out because the circuit may have 5-volt logic circuits. See Figures 3-3 and 3-4.

Again, a Word of Caution

When working with battery connectors, you must keep track of your operating volt- age. If we are not careful, we can mistakenly apply the wrong voltage to the circuit. See Figure 3-4, which shows two different battery voltage sources that have the same connector.

FIGURE 3-3 A 9-volt battery holder with tabs to solder wires to, and a rechargeable 9-volt battery.

FIGURE 3-4 A 3-volt battery holder, a battery connector where the red lead is positive and the black lead is negative, and a 9-volt battery.

Because AA cell battery holders use the same battery connector as a 9-volt bat- tery, we can inadvertently apply the wrong supply voltage to a circuit, which can destroy some of the electronic components. For example, if you are building a 3-volt (digital) logic circuit, then you can use the 3-volt battery holder with 2 AA cells and the battery connector in Figure 3-4. But it is possible, by accident, that the battery connector is snapped onto a 9-volt battery, which will most likely destroy the 3-volt logic circuit. So, we just need to be careful by labeling the battery connector with the operating voltage.

Expected Battery Capacity

You can use a low self-discharge 9-volt NiMh battery in many electronic circuits. Just be mindful that they have about 200 milliamp-hour capacity, which is about 10 per- cent of the AA cells that have generally about a 2000 millamp-hour rating. For proj- ects draining less than 50 mA, the rechargeable 9-volt NiMh battery will work fine. Note that most 9-volt NiMh actually deliver about 9 volts (e.g., 7 cells at 1.25 volts each yield 8.75 volts).

Safety Considerations

If you are using rechargeable batteries, be sure to use only the chargers made for these types of batteries. For example, you can usually purchase a charger along with a set of rechargeable batteries.

Battery chargers are not to be used with primary batteries such as alkaline, car- bon zinc, manganese, and especially lithium cells. For safe practice, do not ever charge any of these batteries!

We will now examine voltmeters, which can confirm the power or battery supply is providing the expected voltage. In particular we will be looking into digital volt- meters (DVMs), which can test voltages from batteries or DC power supplies.

Survey of Digital Voltmeters

The most basic and useful troubleshooting equipment that you can purchase is a volt-ohm-milliamp meter to test for voltage (volts), resistance (ohms), and current (milliamps). Today they are more widely known as digital voltmeters, or DVMs.

One of the basic rules in troubleshooting circuits is to check the power supply with a voltmeter. For example, when a circuit does not turn on, the first thing to check is whether the circuit is receiving the correct voltage. If the voltage is not pres- ent, sometimes it’s important to make sure that the power switch is turned on, or the wires from the power source are connected to the circuit.

The DVM can measure not only the voltage at the circuit, but also confirm with its ohm meter function that a switch or wire has continuity or low resistance.

For now, let’s take a look at some digital voltmeters, which are generally priced from about $6 for the basic ones to $150 for more precision measurements. We will see that because the DVM’s performance-to-price ratio has increased steadily over the years, you can invest in a very versatile DVM that measures not only the voltage, resistance, and current, but also measures characteristics in transistorsdiodes, and capacitors.

With the DVM, we can test the battery voltage, and in some cases, you can pur- chase an inexpensive analog volt-ohm-milliamp meter that tests 1.5-volt and 9-volt batteries.

When troubleshooting a circuit, sometimes it helps to monitor voltage and cur- rent at the same time, or monitor two different voltage points. Because the prices for certain DVMs are relatively inexpensive, you can obtain two or three of these with- out breaking your budget.

Let’s take a look at two similar-looking DVMs in Figure 3-5.

FIGURE 3-5  Two similar DVMs but with vastly different performance levels.

The Lexon (on the left) or the equivalent DT832 DVM (not shown) is rather inexpensive ( approximately $5 to $17 as of 2019) and is good for many general mea- surements. If you compare the Lexon with the Extech, you will notice that the resis- tance range is limited to 2 million ohms (2000k), whereas the Extech goes up to 2000 million ohms (2000M). Another major difference is in their AC voltage range, which is much more extended down to 200 millivolts AC full scale for the Extech compared to 200 volts AC for the Lexon.

Also, if you notice in Figure 3-5, the Extech will measure capacitors, Cx, (up to 20 uf) and an AC signal’s frequency (Hz) up to 200 kHz, whereas the Lexon cannot do any of these measurements.

Note that both DVMs in Figure 3-5 require you to set the measurement range manually. For example, if you are measuring 10 volts DC, you should set the selector knob to “20.” Or if you are measuring a 1 million ohm resistor (1 mega ohm), the selector should be turned to “2000” as in 2000 kilo-ohm since 1 million ohms = 1000 kilo-ohms.

For $20 to $40, you can buy auto-ranging DVMs. See Figure 3-6. With these types of DVMs, the range is automatically set for highest accuracy by the DVM. However, you normally need to wait an extra second or two for the DVM to find the correct range.

FIGURE 3-6 Two auto ranging DVMs by the Extech MN26 and General Tools and Instruments model TS04. Note: Extech replaced the MN26 with the MN36 DVM.

In general, most auto ranging DVMs have excellent measuring capabilities for voltage, current, and resistance. The Extech MN26 can also measure frequency (e.g., 1 Hz to 9.99 MHz) of an AC signal and capacitance (approximately 2 pf to 200 μf), along with temperature when a special probe is used.

The General Tools and Instrument DVM will measure voltage, current, resis- tance, and it has two or three other features. These are

•     Battery tester for 1.5-volt and 9-volt batteries

•     NCV detector that senses live AC power lines or wires by putting the DVM close by the outlets or walls

•     Auto shutoff, which is very useful since some of us forget to turn off our DVMs

•     Bluetooth capability to send measurement data to an iPhone or Android smartphone

Still in the $20 to $40 range, you can get larger-sized, manual-range DVMs. See Figure 3-7.

These DVMs are larger and are easier to read and have good overall measure- ment ranges. However, they do not have as much sensitivity on measuring current (e.g., 2 milliamps versus 200 micro amps for the inexpensive Lexon in Figure 3-5). But if you recall, it’s fine to have a second DVM, which can be something like the Lexon if you need to make more sensitive DC current measurements.

FIGURE 3-7 Two larger-sized DVMs, the DT-9205B and UA890C.

Again, note that the UA890C has the auto-shutoff feature, while this version of the DT-9205B does not. When I searched the web, I did find a “Best” DT-9205M that has “Auto Power Off ”to save on batteries.

Figure 3-8 shows a comparison with another equivalent Lexon digital voltmeter with the DT-9205B.

The less than $10 Cen-Tech DVM has a slide switch for turning it on or off. It does have a battery tester mode for 1.5-volt and 9-volt batteries that the Lexon in Figure 3-5 does not. Note that it too can measure DC current down to 200 micro amps full scale.

There are more high-end DVMs that generally cost $80 to $150 such as the ones made by Keysight (formerly Agilent or Hewlett Packard) and by Fluke. See Figure 3-9.

Both DVMs have the “True RMS” measurement system to accurately character- ize AC signals.

On the Keysight DVM, there is a U1231A version that does not include measur- ing current. I would advise to buy a DVM that includes the three basic measure- ments: voltage, resistance, and current.

FIGURE 3-8  A comparison of a Harbor Freight Cen-Tech DVM with the DT-9205B.

FIGURE 3-9     Keysight U1232A and Fluke 174 precision auto-ranging DVMs.

This finishes the survey on different DVMs that you will be using for trouble- shooting. What I have found is that usually a basic DVM like the Cen-Tech in Figure 3-8 will work for about 90 percent of the time. Once in a while, you may need a DVM that measures capacitance such as the MN26 Extech in Figure 3-6 because it can measure lower value capacitances. Because the prices have come down quite a bit, the basic DVM (e.g., Cen-Tech) can be bought for about $6, and a versatile auto- ranging meter (e.g., Extech MN36, MN26, or MN26T) is in the $44 range. These meters are a good investment for testing parts of your circuits.

In the next chapter, we will look at some basic electronic parts such as resistors, capacitors, and semiconductors (diodes and transistors), which all can be measured and checked with some of these DVMs.

               CHAPTER 4

Some Basic Electronic


We will now look at two electronic components that are often used in electronics— resistors and capacitors.

Resistors pass electricity with varying amounts of conductivity. For example, a perfect wire has no resistance and has ideal conductivity of electricity. On the other hand, a thin wire has resistance and restricts the flow of the electrons somewhat. The higher resistance results in higher restriction of electron current flow. For example, a 10 million–ohm resistor reduces electron current flow to a “trickle,” but it can be useful in many circuits.

However, we will start first with capacitors since they tend to be the ones that can be installed incorrectly and can cause a circuit to fail or work partially.


A capacitor is made of two plates separated by an insulator. So its (DC) resistance is infinite or very high when measured on an ohm meter. See Figure 4-1, which shows a schematic symbol for a non-polarized capacitor.

However, a capacitor does store an electron charge, which makes it like a battery that can quickly charge and discharge DC voltages. If the voltage is an AC signal, it can restrict or pass AC currents depending on its capacitance value, measured in micro farads, nano farads, or pico farads, which are all a fraction of a farad.

One pico farad 5 1 pf 5 1 farad 3 10–12

One nano farad 5 1 nf 5 1 farad 3 10–9

One micro farad 5 1 μf 5 1 farad 3 10–6


FIGURE 4-1  Schematic symbol for a non-polarized capacitor.

Here are some examples of equivalent capacitances:

•     100,000 pf 5 100 nf 5 0.1 μf

•     10,000 pf 5 10 nf 5 0.01 μf

•     1000 pf 5 1 nf 5 0.001 μf

For many capacitors, including ceramic and film type capacitors, the capacitance value is printed in the form of a first and second significant numerical digit and a third multiplier numerical digit, plus a fourth “digit” that is a letter (e.g., K = 10% or M = 20%) to denote the tolerance or accuracy of the capacitance value. The third digit can also be interpreted as the number of zeros after the first two digits.

In general, the three numerical digits tell us the capacitance value in pico farads, which you can equivalently express in nano or micro farads. See the examples below:

•     103 5 10 3 103 pf 5 10,000 pf5 10 nf 5 0.01 μf

•     185 5 18 3 105 pf 5 1,800,000 pf (note the five zeros after the “18”), which is really more commonly expressed as 1.8 μf

•     101 5 10 3 101 5 10 3 10 5 100 pf and NOT 101 pf

•     270 5 27 3 100 5 27 3 1 5 27 pf and NOT 270 pf. However, sometimes a capacitor will be marked by its literal value, and 270 can equal 270 pf. This is why a DVM with a capacitance measurement feature comes in handy.

Many ceramic and film (e.g., polyester, mylar, polycarbonate, or polypropylene) capacitors include a letter to denote the accuracy or tolerance of the capacitance value. The most common letters used are the following list, and the most common tolerances for capacitors are highlighted in the bold fonts:

•     ± 20%

•     ± 10%

•     ± 5%

•     G 5 ± 2%

•     F 5 ± 1%

•     Z 5 –20% to +80

As an example, here is the capacitance range of a 1000 pf 10 percent capacitor, which is marked 102K:

[1000 pf 2 10% of 1000 pf] to [1000 pf 1 10% of 1000 pf] 5

[1000 pf 2 0.1 3 1000 pf] to [1000 pf 1 0.1 3 1000 pf] 5

[1000 pf 2 100 pf] to [1000 pf 1 100 pf] 5

900 pf to 1100 pf

Therefore, it is rare that the capacitor’s capacitance will be exactly as marked. For example, if you measure an off-the-shelf capacitor marked 2.2 μf, it will most likely read below or above that value on the capacitance meter.

Sometimes the capacitor is marked with a “K” after the third numerical digit, which may be confusing as to what the capacitance value really is. This is because the K may also mean kilo instead of 10 percent tolerance. See Figure 4-2.

FIGURE 4-2 Capacitor symbols with typical markings either 4 consecutive “digits” on the left side, or 3 digits plus the tolerance letter below as shown on the right side.

On the left side the 0.01 μf (10,000 pf), 10 percent tolerance cap marked “103K” may be misinterpreted because K may be (incorrectly, in this case) thought of as equal to 1000. See examples A and B below, which are incorrect or misinterpreted:

103 3 1000 pf 5 103,000 pf 5 0.103 μf

or as:

10,000 pf 3 1000 5 10 million pf 5 10 μf

Again the correct value of a capacitor marked as 103K is a 0.01 μf (or 10 nf) capacitor with 10 percent tolerance.

Generally, for electronics work, a farad is too large a value, and capacitors that have capacitances in the 1-farad range are considered to be “super capacitors.” These super capacitors are used in some power supply filter circuits or as a battery to keep memory circuits working when power is cut off.

In many circuits, the capacitors are non-polarized, which means that you can connect them in any manner without worrying about the polarity of the DC voltage across them. For example, a non-polarized capacitor will work properly whether a negative or positive voltage is applied across it.

One of the major “problems” constructing circuits with capacitors is that these days they look “all the same.” In the past, it was easier to spot a larger value capacitor by just looking at the physical size. See Figure 4-3.

FIGURE 4-3 A smaller dimension 0.1 μf (104) capacitor on the left with large physical size 1 μf (105M = 1 μf at 20 percent tolerance) capacitor on the right side.

As we can see in Figure 4-3, we can easily spot the larger capacitance value capacitor by observing both capacitors’ dimensions.

However, with the advent of newer technology, capacitors of small and large val- ues have “equal” sizes. Let’s look at ceramic capacitors that have essentially the same size for a very wide range of capacitances. See Figure 4-4.

In Figure 4-4, the capacitors from left to right are: 10 pf, 100 pf, 1000 pf, 100,000 pf (0.1 μf), 1,000,000 pf (1 μf), and lastly 10,000,000 pf (10 μf).

Because the capacitors of various values look the same size, I would recommend measuring the capacitance function on an appropriate multimeter. For example, see Figure 4-5.

FIGURE 4-4 Ceramic capacitors ranging from 10 pf to 10,000,000 pf (10 μf) that are essentially the same size.

FIGURE 4-5 Measuring a 1 μf (105M) capacitor with an auto-ranging DVM that reads 0.897 μf.

In Figure 4-5, the 105M (1 μf 20 percent tolerance) capacitor shown on the right side of Figure 4-3 measures 0.897 μf, which is well within the 20 percent tolerance since a 1 μf 20 percent capacitor ranges from 0.8 μf to 1.2 μf. Also note that the capacitor is connected only to the test leads, and not to a circuit. When testing a capacitor, it should be out of the circuit. If you have to measure a capacitor in a cir- cuit, first shut off the circuit by turning off the power supply or source, then remove one lead from the circuit and apply the test leads to the capacitor. This method works reasonably well for capacitor values > 1000 pf. For measuring smaller value capaci- tors < 1000 pf, it’s better to remove the capacitor from the circuit and measure the capacitor with a DVM that has a 2 nf (2000 pf) full scale range such as those shown in Chapter 3, Figure 3-7.

NOTE: Beware of large-value, small-sized capacitors with low working voltages; their capacitance can “depreciate” as more DC voltage is applied across them. For

example, some ceramic capacitors may have half the rated capacitance at close to the maximum rated voltage.

We now turn to polarized capacitors that have positive and negative leads. When a polarized capacitor is in a circuit, the voltage across the positive to negative lead must read a positive value at all times. See Figure 4-6 on the schematic symbol.



FIGURE 4-6 A polarized capacitor’s schematic symbol with a curved or bent negative plate to distinguish it from a non-polarized capacitor symbol (Figure 4-1).

A polarized capacitor is usually reserved for larger values, and polarity must be observed. If the voltage across it is not the correct polarity, the capacitor can start drawing DC current and the capacitance value may drop. Worse yet, if the polarized capacitor is connected incorrectly polarity-wise, it may cause injury. As a safety issue, always check the wiring of a polarized capacitor before turning on the power. If there is a problem, rewire the polarized capacitor.

All polarized capacitors have a maximum voltage rating. In general, use a maxi- mum voltage rating of twice the supply voltage if possible. For example, if you have a circuit that runs on 12 volts DC, the electrolytic capacitors should be rated at 25 volts. This 100 percent safety margin allows safe operation for any tolerance and surge voltages from the power supply. Also, the extra maximum voltage rating nor- mally provides longer service life of the capacitor.

For example, if you have a 9-volt powered circuit and you put a 10-volt electrolytic capacitor across the 9-volt supply, chances are that this 10-volt capacitor will lose capacitance faster over time than if you used a 16-volt or 25-volt electrolytic capacitor. Larger value capacitors from 0.1 μf to > 10,000 μf are generally polarized electro- lytic capacitors. Generally polarized electrolytic capacitors have tolerances of ± 20

percent, or in some cases like the “Z” tolerance of –20 percent to +80 percent.

However, it is possible to purchase non-polarized electrolytic capacitors that range from 0.47 μf to 6800 μf. Non-polarized electrolytic capacitors are also known as bi-polarized or bi-polar electrolytic capacitors.

NOTE: You can make a non-polarized (NP) capacitor by connecting two equal value capacitance polarized capacitors back to back in series with the two negative terminals connected, or with the two positive terminals connected. The final capacitance is half of the value of one capacitor. See Figure 4-7.

For example, if you connect two polarized 100 μf electrolytic capacitors in series back to back, the final capacitance will be half of 100 μf, or 50 μf.

NP                                Polarized

NP                      Non           Polarized                        Non           Polarized

FIGURE 4-7 Schematic symbols on the top row for regular, non-polarized (NP), and polarized capacitors, with examples on the bottom row on how to make a non-polarized capacitor with two polarized capacitors in series.

Again often for electronics work, a farad or 1,000,000 μf is too large a value, and capacitors that have capacitances in the 1-farad range are considered to be “super capacitors.” These super capacitors are used in some power supply filter circuits in car audio systems, or as a battery backup DC voltage source to keep low current memory circuits working when the main power is turned off.

Radial and Axial Electrolytic Capacitors

When the two leads from the capacitor protrude from one side (e.g., bottom side), they are radial leaded types. See Figure 4-8 and note the polarity markings and their working voltage.

FIGURE 4-8 Radial lead capacitors at 33 μf, 470 μf, and 4700 μf, all at 35 volts maximum working voltage. Note the (–) stripes and that the negative leads are shorter than the positive leads.

Again, in an aluminum electrolytic capacitor, the negative lead is denoted by a stripe, usually with a minus sign (–).

When the capacitor has leads from both sides, it is an axial lead capacitor. See Figure 4-9.

FIGURE 4-9  A 100-μf, 16-volt axial lead electrolytic capacitor.

Note in the axial lead capacitor, there is an arrow that points to the negative lead.

Often the negative lead is connected to the aluminum case.

Another way to identify the polarity is that the lead or wire from the insulated side is the positive lead. See Figure 4-10.

FIGURE 4-10 A view of the 100-μf, 16-volt electrolytic capacitor showing its positive lead.

Finally for this section, we will show how to make non-polarized electrolytic capacitors from polarized ones. See Figure 4-11.

FIGURE 4-11  Examples of making non-polarized capacitors via a series connection.

As illustrated in Figure 4-7 schematically, we see in Figure 4-11 that two 3.3-μf electrolytic 25-volt capacitors are soldered in series back to back. In the top pair the two positive (+) leads are soldered. And in the bottom pair, the two negative (–) leads are connected. In both cases the result is a non-polarized capacitor whose capacitance is one-half of 3.3 μf or about 1.65 μf.

To be on the safe side, I would rate the maximum voltage at 25 volts and not 50 volts because there is no guarantee that the voltage is equal across each capacitor.

Measure Twice, Install Once: Erroneously Marked Capacitors

Once in a while, it is possible to run into an incorrectly marked capacitor. Figure 4-12 shows such an example.

As we can see from Figure 4-12, the capacitor is marked 0.68 μf, but measures instead 0.157 μf. Thus, this 0.68-μf capacitor is mismarked or defective. In fact, it may be a 0.15-μf capacitor printed incorrectly.

A Proper Method for Measuring Capacitors

1.            Measure the capacitor with a DVM or capacitance meter when the circuit is shut off and at least one of the capacitor’s leads is completely disconnected from the circuit. The DVM or capacitance meter sends out a signal to the capacitor during its measurement. Any “extraneous signal” added to the

FIGURE 4-12 A ceramic capacitor marked 684 = 0.68 μf reads instead 0.157 μf.

capacitor via being connected to the circuit will give erroneous capacitance readings, or worse, can cause damage to the DVM or capacitance meter.

2.            Make sure to not touch either of the capacitor’s lead with your hands. For example, if you hold the capacitor with both hands to make the measurement, your hands can cause an erroneous measurement. This is true when you are measuring low capacitances such as 10 pf. Use alligator or EZ hook test leads to connect the capacitor to the DVM or capacitance meter.

3.            If you are measuring low-value capacitances < 100 pf, then you must subtract off any residual capacitance reading from the DVM or capacitance meter. For example, take note of the capacitance read when the test capacitor is not connected. Usually, this residual capacitance will be in the < 2-pf range. Connect the capacitor and note the reading, and then subtract the residual capacitance. It’s very much like stepping on a bathroom scale that is not zeroed, such as it showing 5 pounds with no weight on it. You then step on the scale, look at the reading, and to get the true weight, you subtract 5 pounds in this example.


Resistors restrict the flow of electric current. But they are used in many circuits to set up a DC voltage to an electronic component such as a transistor or integrated circuit. In other uses, resistors can be used to set the gain of an amplifier or set the voltage output of a DC power supply.

In terms of troubleshooting, the most common “error” in using resistors is mis- reading the color code on the resistor. In some cases, resistors come prepackaged with the resistance value, in ohms, which can be mismarked.

Back in the days before the Internet and apps, most people had to memorize the resistor color code to read the resistance value of the resistor. Today, one can learn the resistor color code, but it is easier to just measure it with the ohm meter that is in virtually all digital voltmeters. A resistor is measured in ohms and has a “shorthand” symbol using the Greek letter omega (Ω). So whenever you see Ω, it means ohm or ohms.

For example, instead of writing 120 ohms, we can equivalently restate it as 120Ω. Figure 4-13 shows a drawing of a standard four-band (e.g., 5 percent tolerance)


1st      2nd      3rd     4th

FIGURE 4-13 Standard tolerance four-band resistor and its associated schematic diagram symbol.

NOTE: A resistor is a non-polarized electronic component. Put another way, there are no polarities in resistors. Thus, you can connect their two leads to a circuit in any order.

To read the four-band resistor, we look at the first band, where the first band is located closer to the edge of the resistor than the fourth band. We start reading the resistor from first and second bands to get the first two digits, then read the third band to multiply by a number such as 0.01, 0.10, 1, 10, 100, 1000, etc. Sometimes people read the third band to add the number of zeroes after the first two digits.

Since each band is a color including black and white, let’s take a look at the code that is based on colors of the rainbow. The colors below are for the first, second and third band, but not the fourth band. Note that the third band denotes the number of zeroes after the first two digits. So if the third band is red, for example, it is not the number 2, but two zeroes added after the first two digits.

The fourth band represents the tolerance, which is either gold 5 5 percent or silver 5 10 percent. There are no other colors than these two for the fourth band.

For example, a 5600-ohm 5 percent tolerance resistor will show the following colors from first to fourth bands:

Green, blue, red, and gold 0 5 black

1  5 brown

2  5 red

3  5 orange

4  5 yellow

5  5 green

6  5 blue

7  5 violet

8  5 gray

9  5 white

Divide by 10 or multiply by 0.10 5 gold

Divide by 100 or multiply by 0.01 5 silver

When the third band has gold or silver “colors,” these are lower-value resistors (< 10Ω). For example, a 2.7Ω 5 percent resistor is denoted as red, violet, gold, and gold. And a 0.39Ω 5 percent resistor has its first to fourth bands as: orange, white, silver, and gold. See Figure 4-14 for examples of four-band color-coded resistors.

Five percent resistor values are incremented in about every 7 to 10 percent. For example, starting from a 220Ω resistor, the next value is 240Ω. If we want to have a 5 percent resistor that is in between, like 226Ω, there is no such value available because there are only two significant digits and “226” requires three significant digits.

Thus we need a resistor with five bands. The first three bands provide the three significant digits, the fourth band gives the number of zeroes, and the fifth band denotes the tolerance that is either brown for 1 percent tolerance, or red for 2 percent tolerance. More commonly, almost all five-band resistors have accuracy to within 1 percent with a brown fifth band. See Figure 4-15.

FIGURE 4-14  Four-band color-coded eighth-watt, quarter-watt, and half-watt resistors.

NOTE: There are some 2 percent five-band resistors where the fifth band is red.

1st      2nd      3rd     4th         5th

FIGURE 4-15  A five-band precision resistor with its schematic symbol below.

Reading five-band resistors can be confusing because you have to look carefully to find where the first digit is. The first band is always a little bit farther from the edge of the resistor than the fifth band. The fifth band denotes the tolerance rating, which is usually brown for 1 percent. Note in Figure 4-14 the fifth band is almost right on the edge of the resistor, whereas the first band is not. Also notice that the spacing between the first four bands are the same, but the spacing is wider from the fourth to fifth bands.

Because the fifth band and the first band from far away may look the same distance from the end or edge of the resistor, you may have to read the resistor both left to right and right to left. Each reading will often give a different resistance value. So my recom- mendation is to take out your DVM and just measure the precision resistor in the ohm meter mode of the DVM. Figure 4-16 shows 5 percent and 1 percent resistors.

FIGURE 4-16 A 22KΩ, 5 percent resistor on the top and 22.1KΩ, 1 percent resistor on the bottom. Note that the fifth band that denotes tolerance rating is closest to the edge in the 1 percent resistor, and it has wider spacing between the adjacent fourth band.

Precision 1 percent resistors and even more accurate resistors with 0.5, 0.25, and

0.10 percent tolerances will have four numbers printed to determine the resistance value, followed by a letter to provide tolerance information. For example, the toler- ance code is

F 5 1 percent

D 5 0.5 percent

C 5 0.25 percent

B 5 0.1 percent

A 1 percent 4750Ω resistor will read 4751F. Remember that the fourth number denotes the number of zeroes after the first three digits.

Another example having a 0.25 percent 10,000Ω resistor will read 1002B. That’s 100 + 2 zeroes afterward from the first 3 digits. This gives 10,000Ω or 10KΩ, where K is a thousand or 1000. See Figure 4-17.

Using a DVM to Measure Resistance Values

Suppose we have a DVM whose maximum resistance ranges are:

•     200Ω

•     2000Ω

•     20kΩ

•     200kΩ

•     2000kΩ (or 2MΩ)

FIGURE 4-17 From top to bottom: 226Ω 1 percent (F), 1000Ω 0.25 percent (C), and a 1000Ω 0.10 percent (B) precision resistors.

Which setting should we use to measure for the most accuracy for a 390Ω 5 per- cent (orange/white/brown/gold) resistor? The answer is the 2000Ω maximum resis- tance setting.

Before we measure resistance using a DVM, remember the following:

•     Turn off the circuit’s power.

•     Disconnect at least one resistor lead from the circuit.

Do not hold the resistor with the DVMs test probes/wires with both your hands. The resistance from your hands may give an erroneous lower resistance measure- ment. This is common when the resistance measurements are > 100kΩ. As a matter of fact, if you have sweaty hands or if the humidity is high in your area, then measur- ing a high resistance value (e.g., > 2000kΩ) will result in the DVM measuring your hand-to-hand resistance in combination with the resistor. This will result in a lower and inaccurate reading. You can hold one probe and one resistor lead together with one hand. And the insulated portion of the second probe is handled by the other hand while the metallic tip of the second probe touches the second resistor lead. You will now get an accurate reading without the DVM measuring your hand-to-hand resistance.

Now let’s take a look at measuring the 390Ω with all the resistance settings in the DVM. See Figures 4-18 to 4-22.

FIGURE 4-18 The 390Ω resistor shows a “1” and blank digits indicating “out of range” at the DVM’s 200Ω setting.

FIGURE 4-19 At the DVM’s 2000Ω setting we see a reading of 386Ω for the 390Ω resistor.

FIGURE 4-20 With the DVM’s 20kΩ setting, the DVM is displaying in kilo-ohms. The 390Ω resistor is measured as 0.38kΩ = 380Ω. Note that we just lost some accuracy compared to Figure 4-19, which measured 386Ω.

FIGURE 4-21 At the 200kΩ setting, the DVM is “looking” for resistors in the 10kΩ range. So the 390Ω resistor is measured as 0.3kΩ = 300Ω. Note that we are further losing measurement accuracy.

FIGURE 4-22 With the DVM’s resistance set to 2000kΩ or 2MΩ, it is expecting resistance values in the 100kΩ range. And with this setting it cannot resolve measuring resistances below 1000Ω. Any resistor value below 1000Ω, such as 390Ω, will register as zero as shown above.

One quick way to measure resistors without thinking too much is to turn the selector until you see the most significant digits displayed.

Measuring Low Resistance Values

In some cases, you will be measuring resistors that are less than 100Ω, such as 10Ω to 1Ω. What you will find in many DVMs is that the test leads may have resistances in the order of 1Ω. It depends on the length of the test leads, which can be short (< 2 inches) or longer (in the 2-foot range).

To make an accurate reading, touch the two test leads together to measure their resistance. Then measure the resistor, and to get the correct value, subtract the test leads’ resistance. See Figures 4-23 and 4-24.

In Figure 4-23 the test leads’ length is short. Normally, the standard test leads that come with a DVM are longer, and may exhibit resistances greater than 0.2Ω. However, if the test leads use heavy gauge wires, the leads may have < 0.2Ω.

Now we will measure a 2.2Ω resistor. See Figure 4-24.

FIGURE 4-23 Shorting the test leads to read their resistance.

FIGURE 4-24  The combined test leads’ resistance (0.2Ω) with the 2.2Ω resistor is 2.4Ω.

To obtain the correct measurement of the resistor, subtract the test leads’ 0.2Ω from the 2.4Ω measurement in Figure 4-24. The answer is then:

2.4Ω 2 0.2Ω 5 2.2Ω

Note that in Figures 4-23 and 4-24, the test leads are very short. Standard test leads are generally 18 inches or more, which can exhibit more than 0.2Ω.

Some DVMs have a zero calibration mode, which subtracts out the test leads’ resistance. For example, in the Keysight U1232A DVM short or connect the two test leads together until the resistance value settles to a constant value such as 2.2Ω (or some other resistance value depending on the resistance of the test leads), keep the test leads connected or shorted together, then hit once the ΔNull/Recall button to zero out the resistance value of the test leads. You are now ready to measure resis- tances with the DVM calibrated.

NOTE: If you shut off the DVM and later turn it on, you should repeat the calibration procedure.

See Figure 4-25 for a picture of the Keysight U1232A DVM with ΔNull/Recall button.

FIGURE 4-25 Keysight U1232A DVM with the ΔNull/Recall button and to the right of it, the Max Min key.

To calibrate and zero out the test leads’ resistance, first connect or short the two test leads until a resistance reading is stable. Then keep the two leads connected or shorted, and then hit/tap once the ΔNull/Recall button that is to the left of the Max Min key. You should now see zero ohms on the display with the test leads connected or shorted together.

We have now covered resistors and capacitors. In the next chapter we will exam- ine semiconductors such as diodes, rectifiers, and zener diodes.

               CHAPTER 5

Diodes, Rectifiers, and

Zener Diodes

Resistors can work properly without regard to “polarity.” That is, in a circuit a resis- tor can be wired either way. For example, you can desolder the resistor from the circuit and flip the resistor leads and solder them back into the circuit and the circuit should work the same.

In this chapter, we will look at some basic two-terminal semiconductors. These are devices that require proper attention to how their leads are wired because their termi- nal leads have polarities. If you reverse a semiconductor’s lead, chances are that the circuit will not work at all, not work well, or not work the same (e.g., symmetrically).

Diodes and Rectifiers

Devices that conduct electricity from one terminal to another, while not conducting electrical current when the terminals are reversed, are diodes and rectifiers.

Before we see this in action, let’s take a look at them in terms of types, connection terminals (e.g., cathode and anode), and schematic diagram symbols. See Figure 5-1.

The schematic symbol and terminal markings are shown in Figure 5-2.

There are two important characteristics in diodes and rectifiers: peak reverse voltage (PRV) and maximum conduction or forward current (IF).

Semiconductors such as diodes, rectifiers, transistors, and integrated circuits

usually have a specific part number printed on them. This is different than the mark- ings on a capacitor where there is the capacitance in micro farads or pico farads and the voltage rating.

Thus, you will not find a diode or rectifier that is marked “50 volts PRV, 1 Amp.” Instead, you will need to look up (e.g., via Google search or data books) the part number based on your diode’s PRV and maximum current rating. In this example, you will find a 50-volt PRV, 1-amp diode has a part number as 1N4001. If you need


FIGURE 5-1 From left to right, a small signal diode (1N914), a 1 amp rectifier (1N4002), and a 3 amp rectifier (1N5401), all with cathode terminals as marked with stripes or bands.

FIGURE 5-2 The schematic symbol of a diode or rectifier is shown on the top and its associated drawing with its cathode marking band is shown on the bottom.

a 3-amp version at 50 volts, the part number will be 1N5400 (e.g., from On Semiconductor).

A typical small signal diode, such as a 1N914 or 1N4148, has a peak reverse volt- age of about 75 volts to 100 volts and a maximum forward current of about 100 mA. Fortunately, in most low voltage circuits today, we do not worry much about the PRV since the power supply voltages generally work at less than 25 volts. However, even low-voltage circuits can damage small signal diodes if the forward current, IF, is exceeded. For example, if a 1N914 small signal diode is used as a power supply rectifier providing 500 milliamps, then the diode may be damaged by excessive

diode current flowing into a circuit. When the diode is damaged, often it will con- duct electricity both ways or directions like a low-resistance resistor, usually in the order of less than 1Ω.

To “correct” the situation, you can replace the 1N914 or 1N4148 small signal diode with a 1N4002 that is a 1-ampere, 100 PRV rectifier. Table 5-1 shows a short list of common small signal diodes and power supply rectifier/diodes with their PRV and maximum current rating.

TABLE 5-1 Silicon Small Signal Diode Part Numbers

Small Signal DiodesPRV (Peak Reverse Voltage)Maximum Forward Current (IF)
1N914100 volts300 milliamps
1N4148100 volts300 milliamps
1N415375 volts200 milliamps

For large signal diodes or rectifiers, see Table 5-2, where these devices are used primarily for power supplies. Large signal diodes/rectifiers may also be used to pro- tect other devices from high voltages, or to prevent circuits from accidental reverse voltages.

TABLE 5-2 Large Signal Diode/Rectifier Part Numbers

Large Signal Diode/RectifierPRV (Peak Reverse Voltage)Maximum Forward Current
1N400150 volts1 Amp
1N4002100 volts1 Amp
1N4003200 volts1 Amp
1N4004400 volts1 Amp
1N4005600 volts1 Amp
1N4006800 volts1 Amp
1N40071000 volts1 Amp
3 Amp Devices Below  
1N540050 volts3 Amps

(continues on next page)

Large Signal Diode/RectifierPRV (Peak Reverse Voltage)Maximum Forward Current
1N5401100 volts3 Amps
1N5402200 volts3 Amps
1N5403300 volts3 Amps
1N5404400 volts3 Amps
1N5405500 volts3 Amps
1N5406600 volts3 Amps
1N5407800 volts3 Amps
1N54081000 volts3 Amps

Large signal diodes such as the ones listed in Table 5-2 can be read relatively eas- ily because the printing is gray or white on a black or dark epoxy diode body. For glass diodes such as the small signal diodes listed in Table 5-1, or other glass body diodes, the part numbers are imprinted on a shorter body, and the numbers are spanned out over three lines or rows. See Figure 5-3.

FIGURE 5-3 “Short body” small signal diodes, such as 1N914B and 1N4148, with their part numbers divided into a succession of three rows with two characters per row.

To identify these smaller glass diodes, we often need a magnifying glass or loupe to read their part numbers. See Figure 5-4.

FIGURE 5-4  Glass diodes and a 5X magnifying loupe to read their markings.

One of the most common mistakes is putting in the wrong diode due to not reading its part number. Smaller glass diodes can include small signal diodes, but also include other types such as Schottky and Zener types. See Figure 5-5 for their schematic symbols compared to a standard diode.

FIGURE 5-5 Top: A standard diode with a straight bar for its cathode; Middle: A Zener diode with a “bent” cathode; Bottom: A Schottky diode—notice that the cathode resembles an “S” (for Schottky).

The drawings in Figure 5-5 are in the order of commonality. That is, we are more familiar with standard diodes first, then Zener diodes that are used primarily for establishing a specific voltage, and finally Schottky diodes that are used in situations where low forward voltage drop across the anode to cathode is required. We will look further into these two devices later. For now, let’s examine some voltage charac- teristics of standard diodes.

Forward Voltage Across Anode to Cathode and Reverse Voltage Effects

When an ideal diode is in forward conduction where a positive voltage is applied to the anode, this diode will transfer the same voltage to its cathode. Therefore, in an ideal diode there is no voltage drop across the anode to cathode when the ideal diode is conducting current. Put in another way, when an ideal diode is conducting cur- rent, it behaves like a wire. If the voltage is reverse biased where the anode voltage is negative, for example, the ideal diode will not pass any electricity and become an insulator or open circuit. The ideal diode has a forward voltage (VF) equal to zero volts.

However, standard silicon diodes have a forward voltage (VF) in the 0.5-volt to 0.7-volt range when measured across the anode to cathode leads. What this means is that when these common silicon diodes and rectifiers conduct in one direction, they develop a “loss” of about + 0.5 volts to + 0.7 volts across the anode to cathode termi- nals. The diode’s forward conduction happens when a positive voltage is applied to the anode with respect to the cathode voltage.

When a standard diode is reverse biased where the cathode to anode voltage is a positive voltage, there is essentially no current flow through the diode until the reverse bias voltage is equal to or exceeds the peak reverse voltage (PRV). Although essentially no current flows during reverse bias, the diode may leak through a very small amount of current, usually a micro-amp or less.

For example, a 50-volt PRV diode (e.g., 1N4001) will exhibit a very high resis- tance when the voltage is less than or equal to +50 volts across the cathode to anode leads. Beyond +50 volts, the 50-volt PRV diode will start conducting current.

When they are not conducting due to reverse voltage applied to the anode and cathode, the resistance is in the order of a million ohms. This reverse bias voltage occurs when the positive voltage is applied to the cathode with respect to the anode voltage, or put in another way, when the negative voltage is applied to the anode with respect to the cathode.

To illustrate forward conduction and reverse voltage non-conduction of a diode, see Figures 5-6 and 5-7.

FIGURE 5-6 Positive terminal of battery is connected to the diode’s anode to enable the diode into forward bias conduction that supplies current to the lit lamp.

In Figure 5-6, we see that a 1N5401 (3 amp) diode supplies current into the lamp. Although a small signal 1N914 diode could have been used, the lamp’s current would have been close to the maximum current rating of the small signal diode. It’s always good to play it safe and use a higher current rating diode. Of course, the

1-amp 1N4002 device would have worked as well.

Now let’s look at what happens when the diode is reversed such that the battery’s positive terminal connects to the diode’s cathode lead. See Figure 5-7.

FIGURE 5-7 Diode connected in reversed bias mode with the battery’s positive terminal connected to the diode’s cathode. No current is flowing and the lamp is off.

Testing Diodes and Rectifiers with Digital and Analog Volt Meters

With digital voltmeters (DVMs), we can determine whether a diode works or not by using the “diode” test feature that is available in most DVMs. See Figure 5-8.

FIGURE 5-8 A typical DVM with its switch set to the “diode” test function that is represented by the schematic symbol of a diode.

We now will measure the forward bias voltage (VF) across the anode to cathode leads of two diodes, a small signal diode 1N914, and a large signal rectifier 1N4002 that has a 1-amp rating. The red positive test lead is connected to the anode and the black negative test lead is connected to the cathode. See Figure 5-9.

In general, higher current diodes or rectifiers will have lower forward voltages than a small signal version. One reason is that the higher current diodes and recti- fiers must have very low equivalent series resistance. This is because when larger currents are flowing through a diode, any lossy resistance internal to the diode or rectifier will add as a forward voltage drop across the anode to cathode leads.

A bad diode will often have shorted out and exhibit the characteristics of a wire, which will conduct both ways. Also, when tested as shown in Figure 5-9, the result- ing voltage will be < 10 mV or 0.010 volt instead of the voltage in the 500 mV to 700 mV range for a standard silicon diode.

When the test leads are reversed where the black test leads are connected to the anodes, and the red positive leads are connected to the cathodes, the DVMs will read a “1” and blank digits to show an “out of range” reading. This out-of-range reading means that the diodes connected this way look like a very high resistance resistor or an open circuit. See Figure 5-10.

FIGURE 5-9 With the DVM’s test leads connected to the anodes via the red positive leads, and the cathodes connected to the black negative leads, we see that the forward voltage (VF) for the 1N914 is +0.637 volt, and the 1N4002 has voltage of +0.593 volt.

FIGURE 5-10 With the red test leads connected to the cathodes (e.g., striped side), the DVMs show an out of range display with only a “1” with blank digits elsewhere. This means that the diodes connected this way are like open circuits or insulators.

NOTE: With most DVMs, the red test lead is like supplying electric current from the positive terminal of a battery. However, this is not always the case with some analog meters.

Suppose we now try to test diodes with the DVM configured as an ohm meter? What we would expect is that in the forward conduction mode, the DVM’s ohm meter will display some resistance value. However, this may not be the case at all. See Figure 5-11 where we compare measuring a rectifier with an analog VOM (volt- ohm-milliamp meter) and a DVM.

FIGURE 5-11 1N4002 rectifiers being tested on the left with an analog ohm meter that shows about 70Ω. The DVM on the right side shows an out-of-range condition signifying an open circuit, which is erroneous. Black leads of both meters are on the cathodes.

NOTE: The Triplett 310 analog ohm meter shows a 70Ω resistance with the diode at forward conduction when the positive test lead is connected to the anode. However, the actual resistance is lower because the diode has a forward voltage drop (e.g., 0.6 volt) that results in an inaccurate resistance reading. Usually, a reading in the forward conduction mode of less than 150Ω but more than 2Ω indicates that the diode is OK.

Many DVMs may not be able to test diodes in their ohm meter mode. Virtually all analog VOM meters will be able to test conduction of diodes in one direction while showing very high resistance in the reverse direction.

In Figure 5-11, the red positive test lead of the analog meter is connected to the anode of the diode-rectifier for testing forward conduction resistance. If the leads were reversed where the black lead is connected to the anode and the red lead is con- nected to the cathode, the meter’s needle will not move and stay on the side of the top scale, which would show infinite resistance.

With analog meters in the ohm meter mode, the red test lead does not always relate to the positive terminal of a battery. For those older meters made in United States, such as the Simpson 260 or the Triplett 310, 620, and 630 series, these analog VOMs have their positive test leads relating to the positive terminal of a battery.

However, some but not all analog VOMs elsewhere have their polarities reversed. That is, the black negative test lead is connected to the positive terminal of a battery. See Figure 5-12.

FIGURE 5-12 A “Micronta” analog VOM tested shows forward conduction (11.5Ω on top scale) with the negative black test lead connected to the rectifier’s anode.

Notice that the diode’s forward conduction occurs when Micronta meter’s nega- tive test lead is connected to the anode. This is different from Figure 5-11 where the

Triplett ohm meter shows forward conduction when its positive red test lead is con- nected to the anode of the rectifier.

Again in Figure 5-12, if the leads were reversed such that the red positive test lead is connected to the anode and the black test lead of the Micronta meter is con- nected to the cathode, then the diode will show essentially infinite resistance.

Testing a diode is normally done with analog VOM set to Rx1 (Ohms x 1) or Rx10 (Ohms x 10). Also, the actual resistance value for the diode in the forward conduction mode may vary from one analog voltmeter to another. The reason for this variation is because each meter may supply different test current or voltage to the diode. Usually, if the diode’s forward conduction resistance is less than 150Ω with Rx1 selected, the diode is OK.

Usually, a bad diode will short out and the resistance will read close to 0Ω. If you see this, then replace the diode. But also inspect or test other electronic components connected to the diode.

NOTE: Always test diodes with the power shut off and disconnect one lead of the diode from the circuit. If a diode is tested “in the circuit,” then an erroneous reading may occur. The reason is that other parts of the circuit connected to the diode may add a parallel resistance across the anode and cathode leads of the diode. This parallel resistance can cause the diode’s measurement in the reverse bias mode to show an unusually low resistance.

Schottky Diodes

Recall the schematic symbol for a Schottky diode has the cathode resembling an “S.” Schottky diodes have lower voltage drop across their anode to cathode terminals compared to standard silicon diodes. With this lower voltage drop, Schottky diodes work well for lower voltage power supplies as to not “take away” some of the raw supply voltage. For example, if the raw supply is furnishing only 1.5 volts, a standard diode’s forward voltage drop (VF) is 0.55 volt that will subtract from 1.5 volts. As a result, about 0.95 volt will be at the output of the standard diode. A Schottky diode has a lower forward voltage drop, such as VF 5 0.15 volt, and thus the output voltage will be 1.5 volts – 0.15 volt 5 1.35 volt, a vast improvement over the 0.95 volt output

from the standard silicon diode.

In general, you can replace standard silicon diodes with Schottky diodes, but be aware that the PRV rating may be lower. Schottky diodes are commonly used for switching power supplies that typically provide voltages below 24 volts DC but at very high currents.

For example, Table 5-3 shows the voltage ratings for the following 3-amp Schottky diodes. Note the lower PRV ratings compared to the 1N5400 series devices.

TABLE 5-3 Schottky Diodes Characteristics for the 1N5820 Series 3-Amp Devices

Schottky DiodesPRV (Peak Reverse Voltage)Maximum Forward Current, IF
1N582020 volts3 amps
1N582130 volts3 amps
1N582240 volts3 amps

In contrast, the standard silicon 3-amp diode 1N5408 has a 1000 volt PRV.

Now let’s take a look at the forward conduction voltage drops of Schottky diodes.

See Figures 5-13 and 5-14.

FIGURE 5-13 Testing for VF a Schottky diode (1N5822) on the left side, and a standard silicon diode (1N5402) on the right. Black leads of both meters are on the cathodes.

Note that the Schottky and silicon diodes look physically identical but the Schottky has a 0.153 forward voltage (VF) compared to a larger 0.533 voltage drop for the standard version.

Now let’s turn to small signal diodes. See Figure 5-14 for measurements of VF.

FIGURE 5-14 A Schottky small signal diode (1N5711) on the left and a standard small signal 1N914 diode on the right. Black leads of both meters are on the cathodes.

Note that the Schottky diode here has a 0.346-volt drop versus the 0.636-volt drop for the standard silicon 1N914 small signal diode.

In small signal circuits such as those with radio frequencies in a crystal radio, a standard silicon diode like the 1N914 or 1N4148 is unsuitable as the crystal detector because the radio frequency signals are weak, typically < 300 mV. A small signal Schottky is better but will still not pass through weaker radio signals in the crystal radio.

A better detector for a crystal radio is the germanium diode. It is based on very old technology (before 1950) and the germanium diode has a lower forward conduc- tion voltage (VF), typically less than 300 millivolts. See Figure 5-15.

As we can see here, the germanium diode has a 0.233-volt drop versus the 0.638

volt in the silicon diode. Note that the Schottky diode measures 0.346 volt in Figure 5-14.

Also, beware of vendors selling “germanium” diodes that are of the size of a 1N914. These are unlikely to be (e.g., 1N34) germanium diodes. It should be noted that germanium diodes are generally packaged in larger (physical) sizes than small signal silicon diodes. However, a few vendors on the web may try to sell their germa- nium (e.g., 1N34) type diodes that are silicon and have the same larger size as a germanium diode. Thus, beware and (to be certain) always measure the forward diode voltage as shown in Figure 5-15.

FIGURE 5-15 A germanium 1N34 diode on the left and a silicon 1N914 diode on the right. Note that the physical size of the germanium diode is much larger than the silicon device. Black leads on both meters are on the cathodes.

For high frequency or radio frequency circuits (e.g., whose frequencies > 400 kHz), small signal diodes also require very low internal diode capacitance between the anode and cathode leads. The reason is that if there is a large capacitance across the anode and cathode, this large capacitance will cause signals to leak through the diode and bypass the diode as a signal detector.

For example, if we look back at Figure 5-14, then the 3-amp Schottky has a 0.152- volt voltage drop, even lower than the 1N34 germanium diode at 0.233 volt. So why not use the Schottky diode instead? The reason is that the 1N5822 Schottky rectifier has hundreds of pico farads of internal capacitance across the anode and cathode leads. This will cause the radio signal to bypass the 1N5822 rectifier, and thus the radio signal will not be demodulated correctly. However, if a radio signal’s frequency is in the 50 kHz or lower range, then the 1N5822 Schottky rectifier may just work.

A Brief Look at Zener Diodes

Zener diodes are very similar to standard silicon diodes in that they can pass current in one direction such as having a positive voltage applied to the anode. But this is not

how Zener diodes are used. Instead, Zener diodes have a set or predetermined and accurate PRV (peak reverse voltage). This type of PRV in Zener diodes is known as the Zener voltage.

Some Zener voltages are:

•     3.3 volts

•     3.9 volts

•     4.3 volts

•     4.7 volts

•     5.1 volts

•     5.6 volts

•     6.2 volts

•     6.8 volts

•     8.2 volts

•     12 volts

•     15 volts

A Zener diode is also rated in wattage, which allows you to determine the maxi- mum current flowing into it. Some of the common wattage ratings are 400 milli- watts, 500 milliwatts, 1 watt, and 5 watt.

The maximum Zener diode current, IZener 5 Wattage rating/Zener voltage

For example, if we have a 5.6-volt, 5-watt Zener diode (1N5339B), the maximum current is:

IZener 5 Wattage rating/Zener voltage

IZener 5 (5/5.6) amp 5 0.89 amp 5 890 mA

In another example, suppose we have a lower wattage Zener diode such as the 1N5237B, rated at 500 mW (0.50 watt) and 8.2 volts.

IZener 5 (0.50/8.2) amp 5 0.061 amp 5 61 mA

One thing to keep in mind as the Zener voltage is increased with the same watt- age, the Zener diode’s maximum current goes down in proportion. But if the Zener voltage is decreased, then the Zener diode’s maximum current will increase given the same wattage rating.

TABLE 5-4  Example Maximum Zener Diode Current Based on Wattage and Zener Voltage

Zener DiodesWattageZener VoltageMaximum Zener Current IZener
1N47371 watt7.5 volts133 mA
1N47441 watt15 volts66 mA
1N47511 watt30 volts33 mA

From Table 5-4 we first look at the 1N4744 Zener diode at 15 volts with a Zener current of 66 mA. If we decrease the Zener voltage by 2, which is a 7.5-volt Zener diode (1N4737), we have twice the Zener current at 133 mA. And increasing the 15-volt Zener voltage to 30 volts (1N4751) leads to half the Zener current at 33 mA. In general, you should try to keep bias on the Zener diode to not more than 75 percent of the maximum Zener current. A good starting point is running the Zener diode at somewhere in the 15 percent to 50 percent maximum of IZener. For example,

the 1N4744 device has a 66 mA maximum Zener current, and you can bias it at

50% or 33 mA.

Now let’s take a look at how various Zener diodes are labeled. See Figure 5-16.

FIGURE 5-16 An example of a 0.50-watt 1N5237B Zener diode, and of a 5-watt 1N5339B Zener diode with their associated labeling.

Note in Figure 5-16 that the markings can span two or three lines for identifying the Zener part number. In the example of the 1N5237B, actually the first two lines (rows) will give your correct part number. You can usually ignore the suffix “B.”

For the 1N5339B, we see in Figure 5-16 that the part number is spanned over three lines (rows) to identify the Zener diode.

Just a reminder, a positive voltage greater than the Zener voltage will be con- nected to the cathode of the Zener diode. But the Zener current will be limited via resistor or a current limiting circuit.

So a pure voltage source such as a power supply or battery that has a higher volt- age than the Zener voltage is NEVER connected across the Zener diode.

Now let’s look at two examples of Zener diodes in simple circuits. See Figures 5-17 and 5-18. Figure 5-17 shows that the red lead is on the anode.

FIGURE 5-17 A 4.3-volt 1N749A, 400 mW Zener diode is connected in series with a resistor to drop or subtract 4.3 volts from the 9-volt battery.

In Figure 5-17 we see the 9-volt battery’s positive terminal connect to the cath- ode of the 4.3-volt Zener diode. Had this been a regular diode such as a 1N914, we would have a reverse bias condition and the voltmeter would have shown 0 volts. But because this is a Zener diode, it “breaks down” at about 4.3 volts and forms a 4.3 positive voltage from the cathode to the anode. This cathode to anode voltage, which is the Zener voltage, subtracts from the battery voltage when the voltage is measured from the anode of the Zener diode and the negative terminal of the battery. The resistor value shown is 1000Ω, and 4.82 volts is measured on the voltmeter. Note that 9 volts – 4.3 volts 5 4.7 volts, which is approximately 4.82 volts. If the resistor’s value was changed from 1000Ω to 2000Ω or to 510Ω, the measured voltage would still be about 4.82 volts.

Let’s take a quick look at what happens if a Zener diode is accidentally reversed in wiring.

See Figure 5-18 for a Zener diode’s forward voltage, which is similar to a regular silicon diode.

FIGURE 5-18 A 4.3-volt 1N749A, 400 mW Zener diode on the left and a standard silicon diode, 1N914 on the right. Both black leads are on the cathodes.

As shown in Figure 5-18, both diodes are forward biased by having the red posi- tive test leads connected to the anode of the Zener diode and standard diode. The Zener diode shows 0.789 volt while the standard small signal diode (1N914) shows 0.633 volt.

Thus, if you reverse a Zener diode’s connections you will get a voltage lower than the Zener diode voltage. For example, the 4.3-volt Zener diode provides about + 4.3 volts (across cathode and anode) when connected correctly, but will give only about

+ 0.789 volt when the leads are reversed.

Now let’s see if there is a better way to use the Zener diode. Figure 5-17 shows a series connection method of using a Zener diode. But it is not necessarily the best method to provide a stable voltage, or in an accident, a safe voltage. For example, sup- pose the 4.82-volt output voltage at the resistor is connected to a 5-volt logic circuit. If the Zener diode is inadvertently reversed in connection where the positive terminal is connected to the anode, the resistor will instead have 9 volts – 0.7 volt 5 8.3 volts.

If 8.3 volts is applied to a 5-volt logic circuit, most likely there will be damage. Also, if the 9-volt battery drops in voltage, the voltage at the anode of the Zener will drop as well below 5 volts. For example, suppose the 9-volt battery drains down to 8 volts. Then the output voltage in Figure 5-17 will be 8 volts – 4.3 volts 5 3.7 volts. In the next example we will show a different circuit known as a shunt regulator that will maintain essentially constant voltage as long as the battery or power supply voltage is greater than the Zener diode voltage.

A better circuit is shown in Figure 5-19. Should the Zener diode be inadvertently connected in reverse or if the Zener diode shorts out, the voltage will be 0.7 volts or close to zero volts. This will be below the nominal 5 volts and will provide an “under voltage” that does not harm a circuit.

FIGURE 5-19 A 5.1-volt 1N751 400 mW Zener diode configured as a “shunt” regulator circuit that provides 5.13 volts as shown in the DVM. The red lead is on the cathode.

In Figure 5-19, the output voltage is taken across cathode to anode of the Zener diode. This output voltage is the Zener voltage. The series resistor (e.g., 1000Ω) is connected to the battery’s positive terminal and furnishes current to the Zener diode’s cathode. A positive voltage is provided at the cathode with the anode con- nected to the battery’s negative terminal. Figure 5-19 shows a shunt regulator circuit because the Zener diode is “shunted” across the output terminals.

Suppose the Zener diode is accidentally reversed. This accident results in the anode connected to the resistor and cathode connected to the battery’s negative ter-

minal. The voltage will be the forward voltage of the diode, which is about + 0.7 volt. So, if a 5-volt logic circuit is connected to this shunt regulator circuit, there will be no damage to the circuit due to over voltage. And if the diode should short out, the volt- age will be close to zero volts, which again will not damage the 5-volt logic circuit.

Some General Rules About Diodes

•     They should not be wired in parallel for greater current capability because diodes bought off the shelf are not matched. If you measure the forward diode voltage from anode to cathode and the diodes are “matched” within 5 mV (e.g., Diode #1 5 0.608 volt, Diode # 2 5 0.603 volt), then they can be paralleled in close proximity to each other.

•     If you buy the diodes in “tape” form, you have a much better chance of having matched diodes. The matched diodes can be used in circuits requiring nearly identical forward voltage characteristics and is not limited to paralleling diodes. See Figure 5-20 and be sure to double-check the diode’s forward voltages (VF) with a DVM. Both black leads are on the cathodes.

FIGURE 5-20 Two 1N914 diodes measured from a “tape” of 4 successive diodes with very close matching of VF within 5 mV. In this example the two diodes’ forward voltage readings of 0.641 volt and 0.642 volt are only 1 mV apart from each other.

So this ends a lengthier than normal chapter with introductions on standard, Schottky, and Zener diodes.

We will take a “breather” with a slightly shorter Chapter 6 that deals with light- emitting diodes (LEDs).

                CHAPTER 6

Light-Emitting Diodes

In this chapter we will explore light-emitting diodes (LEDs) of different emitted wavelengths (e.g., red, green, and blue), and how their “turn-on” voltages (VF) are related. We will also discuss common problems that can cause an LED to fail, such as over current or “excessive” reverse voltage.

The LED’s Light Output

Light-emitting diodes are measured for luminous intensity (light output) in milli- candelas (mcd) at a particular LED current. There are various descriptions of LEDs such as “high efficiency,” “high intensity,” or “high performance.” However, the most important specification to look for is the light output for a given LED current, which is usually done at 10 mA or 20 mA.

With the exception of using older LEDs that give out about < 10 milli-candelas at 10 mA current, which are suitable for indicator lamps such as power-on lights, you will find that most LEDs are much brighter.

For example, typical LEDs will provide over 1000 mcd at 20 mA. In keychain LED flashlights or multiple LED flashlights, the rating is in the order of > 10,000 mcd at 20 mA. The extra brightness works well in products such as light cubes, pho- tographers’ studio or portable lighting, emergency lights, home light bulbs and light fixtures, or automobile headlights.


Here are some basic rules of thumb on LEDs:

•     For indicator lamps, generally with today’s LEDs of any color—red, green, yellow, blue, or white—the LED current can be in the 1 mA to 5 mA range. With very high-brightness LEDs such as a 20,000 mcd unit, the drive current can be as low as 0.10 mA or 0.25 mA.

•     If you are using LEDs to generate intense lighting such as making a flashlight or a light cube, you should not exceed 20 mA per LED. Otherwise, the LED may take in excessive current, which causes it to burn out.

•     For those situations where intense LED light is generated, do not look directly into the LED light source. You can damage your eyesight.

•     The light output is generally proportional to drive current. For example, if you are using an LED that is providing 20,000 mcd at 20 mA. Suppose you want to turn it down so that it can be used as an indicator lamp. If the LED drive current is reduced by 200 times from 20 mA to 0.10 mA, the LED will output 100 mcd, which is still pretty bright for an indicator lamp.

NOTE: 20,000 mcd divided by 200 = 100 mcd.

Standard LEDs come in 3mm, 5mm, and 10mm sizes. In this chapter, we will be looking at the common 5mm types that have 100 mil (0.1 inch) lead spacing.

Now, let’s take a look at some LEDs.

Figure 6-1 shows a red, green, and yellow 5mm size LED from left to right. Notice that each of the LEDs comes with a longer lead, which is identified as the anode. Each of these LEDs will turn on at less than 2 volts DC, typically about 1.7 volts. Also, they have a “frosted” lens that allows for a wider viewing angle and is easier to look at because the light is more diffused.

FIGURE 6-1 Red, green, and yellow LEDs give out “diffused” light due to their encapsulations with their associated colors.

In Figure 6-2, the blue and white LEDs turn on at a higher voltage, around 2.5 volts to 3.0 volts DC. The blue and red LEDs have a frosted lens that allows for a wider viewing angle. The middle clear lens, white LED is made for maximum bright- ness but over a narrower viewing angle. Clear lens LEDs are used for flashlights or for efficient projection/illumination of a smaller area.

FIGURE 6-2 A blue, white, and again a red LED (shown from left to right) that are 5mm in size.

Again, the spacing between the two leads in 5mm LEDs is 0.1 inch. To identify the LED’s anode and cathode leads see Figure 6-3.

FIGURE 6-3 Identifying the anode and cathode leads by length or by locating the “cup” that is connected to the cathode lead. From the side of the LED, the narrower lead as shown is the anode.

We now turn to the LED’s schematic symbol. See Figure 6-4.

Anode            Cathode

FIGURE 6-4 An illustration identifying the anode and cathode via the lead lengths (e.g., longer lead is the anode) and its schematic symbol with two outward arrows signifying light emission.

It should be noted that the schematic symbol is often drawn vertically. See Figure 6-5.

FIGURE 6-5 Often the LED symbol is drawn vertically with the cathode grounded and the anode connected via a resistor to a positive power supply.

Before we look at lighting up LEDs, there is one “trick” to always identify a white light LED. See Figures 6-6 and 6-7.

FIGURE 6-6 A red (R), yellow (Y), green (G), or infrared (IR) LED on the left, and a white LED on the right.

In Figure 6-6 you can always identify a white light LED by its yellow phosphor when looking directly into it without power applied to it. If you see “gray,” it is an LED other than white.

Figure 6-7 shows a closer look at the white LED’s yellow phosphor.

FIGURE 6-7 Viewing straight from the top, you will see yellow throughout, which signifies that this is a white LED.

NOTE: A burnt-out white LED may be missing the yellow phosphor.

The yellow phosphor is excited by a blue LED below it, which, when combined, forms a white light. Note that on the color wheel, blue and yellow are at opposite ends. In fact, if your photo has a yellow color cast, you add blue to remove the yellow cast. Conversely, if the photo has a blue cast to it, you add yellow to return the picture to normal colors.

LED “Minimum Turn On” Voltages

In this next section we will measure the turn-on voltages for LEDs. The turn-on volt- ages of particular LEDs are relatively constant, almost regardless of LED current. This means that a slight voltage increase (e.g., +100 mV DC more than the previous voltage) across the anode to cathode leads can result in an increase in LED current by tenfold or more. This is why generally LEDs are driven by a resistor or some type of current limiting device such as a transistor.

When the LED voltage and current characteristics are well known, then some- times a voltage source can be applied to the anode and cathode terminals. For exam- ple, we see this in some three AAA cell multiple LED flashlights. See Figure 6-8. In this specific case, there is somewhere between 3.6 volts to about 4.0 volts directly across the anode and cathode of each of the white LEDs. The minimum turn-on voltage of white LEDs is between 2.5 volts and 3.0 volts. Although there are three 1.5-volt AAA cells in series to provide nominally 4.5 volts, by the time all the LEDs are lit, there is enough current draw to lower the three AAA batteries’ 4.5 volts to

about 4.0 volts or lower (measured 3.93 volts with three alkaline AAA cells when the LEDs are turned on), which is just safe enough to prevent the LEDs from burning out. In this example, the three series connected AAA cells have sufficient internal current limiting resistance. This allows the LEDs to be powered on safely.

NOTE: If you use higher capacity batteries that have lower internal resistance than the AAA batteries such as C or D cells, the LEDs’ life may be shortened significantly due to excessive current. The reason is that the LEDs will essentially receive the full

4.5 volts from the three series connected C or D alkaline cells.

FIGURE 6-8 A three AAA cell 9 LED flashlight, where the LEDs are matched in terms of turn-on voltage.

In Figure 6-8, the three AAA cells are connected in series and provide sufficient internal series resistance for LED current limiting. The nine matched LEDs are con- nected in parallel (e.g., all nine anode leads and all nine cathode leads are connected) that provide equal brightness among each of the LEDs.

If you randomly take nine white LEDs and connect them in parallel, the bright- ness will vary between some of these nine LEDs due to different turn-on voltages.

For long LED life, you will need to drive them with batteries that have some internal resistance. Using three AAA batteries will be fine for driving the LEDs within their current limits. But to reiterate, if you drive the LEDs with three C or three D cells, you can burn out the LEDs. The reason is that the C and D cells have lower internal resistance (than the AAA cells) and can deliver much more current into the LEDs.

Now let’s take a look at LEDs whose turn-on voltages are typically less than 2 volts DC from anode to cathode. See Figure 6-9.

FIGURE 6-9 LED turn-on voltages (VF) of 1.786 v = Red, 1.872 v = Green, and 1.842 v = Yellow. The drive voltage is 9 volts with an 8060Ω 1 percent series resistor for each LED.

See Table 6-1 for color and wavelength related to Figure 6-9.

TABLE 6-1  LED Turn-On Voltages and Their Wavelengths

LEDTurn-on Voltage (VF)Wavelength
Red1.786 v~ 640 nanometers
Yellow1.842 v~ 583 nanometers
Green1.872 v~ 565 nanometers

Notice that the turn-on voltages increase as the wavelengths get shorter. For example, the longer wavelength red LED has 1.786 volts turn-on voltage, whereas the shorter wavelength green LED has a turn-on voltage of 1.872 volts.

See Figure 6-10 for a schematic diagram of the LEDs being lit in Figure 6-9. In this schematic the LEDs have one arrow emanating from them to denote light emis- sion. Generally, whether there is one or two arrows emanating from the diode sym- bol, the meaning is the same—the diode is an LED.


Tekstvak: 2BT1

R1 8060


R2 8060

RL                               RL

R3 8060

9 Volt Battery

+1.786 v

Red LED1

+1.872 v

Tekstvak: 1Tekstvak: 1Tekstvak: 2Green LED2

+1.842 v

Tekstvak: 1Tekstvak: 2Yellow LED3

BL                             BL                               BL

Common Point or Ground

RL = Red Lead of DVM BL = Black Lead of DVM

FIGURE 6-10 Schematic diagram of the red, green and yellow LEDs with their respective forward voltages (VF) at 1.786 v, 1.872 v, and 1.842 v.

The forward LED current (IF) is the battery voltage minus VF divided by the drive resistor, such as R1, R2, or R3.

For example, the current flowing through the red LED is

IF 5 (9 volts – 1.786 volts) / R1 5 (9 volts – 1.786 volts) / 8060Ω

IF 5 7.214 volts / 8060Ω

IF 5 0.895 milliamp

Notice in Figure 6-10, a “ground” or common point terminal has been added that is connected to the negative terminal of the 9-volt battery. Often, the DVM’s negative or black test lead is connected to the ground or common point. Then the DVM’s red test lead is “free” to probe or measure for voltages at different parts of a circuit.

Now let’s take a look at turn-on voltages for blue, white, and red LEDs.

FIGURE 6-11 White (W), blue (B), and red (R) LEDs lit via a 9-volt battery and 8060Ω resistors.

Both white and blue LEDs are very bright, and the blue LED is actually overex- posing the camera’s sensor. See Figure 6-12 for another look at these three LEDs lit.

FIGURE 6-12 The white (W), blue (B), and red (R) LEDs lit.

Now let’s take a look at Table 6-2, a summary of the turn on-voltages for white, blue, and red LEDs.

TABLE 6-2  Turn-On Voltages for White, Blue, and Red LEDs

LEDTurn-on Voltage (VF)Wavelength
White2.67 vNot applicable to white light
Blue2.60 v~ 470 nanometers
Red1.78 v~ 640 nanometers

Note that since a white LED really uses a blue LED inside with a yellow phos- phor, both white and blue LEDs have essentially identical turn-on voltages, 2.67 volts and 2.60 volts.

A schematic diagram is shown in Figure 6-13.



R1 8060


R2 8060

RL                               RL

R3 8060

9 Volt Battery

+2.67 v

+2.60 v

Tekstvak: 2White

Tekstvak: 1LED1

Tekstvak: 1Blue LED2

+1.78 v

Tekstvak: 2Tekstvak: 1Tekstvak: 2Red LED3

BL                             BL                               BL

Common Point or Ground

RL = Red Lead of DVM BL = Black Lead of DVM

FIGURE 6-13 A schematic diagram showing the turn-on voltages (VF) measured for white, blue, and red LEDs.

Again, we can determine the forward LED current (IF) by the following: IF 5 (supply voltage – VF) / drive resistance

For the blue LED, we have: IF 5 (9 volts – 2.60 volts) / R2

IF 5 (9 volts – 2.60 volts) / 8060Ω 5 6.40 volts / 8060Ω

IF 5 0.794 milliamp

Other Types of Green LEDs

In Table 6-1, we found that the green and yellow LEDs had very close turn-on volt- ages at 1.872 and 1.842 volts. And if you look carefully at these types of green and yellow LEDs, we notice that the green LED gives closer to a green-yellow color instead of a darker or deeper green hue without much yellow. (See Figure 6-9 for an example). It turns out that there are green LEDs that are closer to green-blue in tint, which is a “truer” rendition of what we think of as green color.

Also, there are LEDs that give out more light such as the green 1-watt LED in Figure 6-14.

FIGURE 6-14 A 1-watt LED with its anode marking as shown with the dot on one lead. At 1 watt, the maximum current for this LED is about 300 mA. The LED has to be heat- sinked properly by soldering the two leads to large areas of copper that dissipates the heat away.

Let’s take a look at the turn-on voltages between a standard green-yellow LED and a green-blue LED. See Figure 6-15.

FIGURE 6-15 The 2.58-volt turn-on voltage for the (phosphor) green-blue (G-B) LED is closer to the turn-on voltage of LEDs that are white (2.67 v) or blue (2.60 v). Note that a standard green LED, which gives green-yellow (G-Y) light has a turn on voltage of 1.88 v.

We can see a difference in colors of the two “green” LEDs in Figure 6-16.

FIGURE 6-16 A shorter wavelength green-blue (528 nm) LED on the left shows a deeper green compared to the longer wavelength green-yellow (565 nm) LED on the right. Note: nm = nanometers.

Although the example in Figure 6-16 shows a 1-watt LED, the green-blue LED is available in the same (5 mm) size as the green-yellow LED.

Problems with Paralleling Two LEDs with Different Turn-On Voltages

In Figures 6-9 to 6-13, 6-15, and 6-16 the LEDs are driven via a series resistor so that the LED current can be controlled more precisely. For example, see Figure 6-17.



R1 8060


R2 8060

RL                               RL

R3 8060

9 Volt Battery

+2.67 v

+2.60 v

Tekstvak: 2White

Tekstvak: 1LED1

Tekstvak: 1Blue LED2

+1.78 v

Tekstvak: 2Tekstvak: 1Tekstvak: 2Red LED3

BL                             BL                               BL

Common Point or Ground

RL = Red Lead of DVM BL = Black Lead of DVM

FIGURE 6-17 These LEDs include series current-limiting resistors R1, R2, and R3.

However, what happens if we only drive 2 paralleled LEDs with just one current- limiting resistor? See Figure 6-18.

FIGURE 6-18 A red LED and a green LED connected in parallel are driven via one series current-limiting resistor. Note that the red LED is much brighter than the green LED.

For a close-up of the two LEDs, see Figure 6-19.

FIGURE 6-19 A close-up view of the two LEDs in parallel that shows the green LED being much dimmer than the red one.

The circuit is shown schematically in Figure 6-20.

The measured voltage across the two LEDs is 1.81 volts. However, from Figure 6-9, the different turn-on voltages with 8060Ω driving resistors resulted in 1.786 volts for red and 1.872 volts for green. Hence, there is about an 86 mV difference with the red LED that turns on with a lower voltage. As shown in Figure 6-20, the measured voltage at the anode is 1.81 volts with a 4300Ω resistor. Note that the red LED’s 1.81-volt VF is still lower than the 1.872-volt turn-on voltage of the green LED

in Figure 6-9. Hence, the red LED will divert or hog most of the current and it will light up brighter than the green LED (see Figure 6-19).


Tekstvak: 19 Volt Battery



Tekstvak: 2+1.81 v


Green LED1

R1 4300

Tekstvak: 1Tekstvak: 2Red LED2

Common Point or Ground

RL = Red Lead of DVM BL = Black Lead of DVM

FIGURE 6-20  Two LEDs are wired in parallel with a single series resistor R1.

In general, if you are going to parallel LEDs together, first they should be of the same color emission. Second, they should be matched, which, for example, can be done by purchasing the LEDs in “tape” form. See Figure 6-21.

FIGURE 6-21 LEDs bought in tape form are generally better matched in terms of turn-on voltages than those bought in a bag or off the shelf.

Protecting LEDs from Damage Due to Reverse Voltage Across the Anode and Cathode

If you are going to connect LEDs to an AC (alternating current) voltage source such as an AC adapter that gives AC voltage (e.g., 6VAC, 9VAC, 12VAC, etc.) you will need to protect the LEDs from damaging reverse voltage. Recall that AC voltages provide positive voltages at one time followed by a negative voltage at another time. For example, a 60 Hz AC voltage source has a repetition rate of (1/60) second or 16.666 millisecond. In the first half of that period, the voltage is positive for 8.333 milliseconds (e.g., half cycle). And in the next 8.333 milliseconds (e.g., the remain-

ing half cycle) the voltage is negative.

LEDs only have a peak reverse “breakdown” voltage of about 5 volts. This means for all practical purposes, they must be protected should AC voltage sources be con- nected to them. Figure 6-22 shows an example with a rectifier (e.g., 1N4004) con- nected in series with the LED.

FIGURE 6-22 A limiting resistor (e.g., 1000Ω) is connected to the anode of the rectifier, and the rectifier’s cathode is connected to the anode of the LED.

A schematic diagram is shown in Figure 6-23 on the left side.

D1 Rectifer                                                   D1 C1 Circuit


1K to 10K

D1                                          C1

D1 1N4004


AC Voltage


Tekstvak: 1Tekstvak: 2Green LED1


220 uf 50 v AC Voltage

3K to 30K

Tekstvak: 1Tekstvak: 2Green LED1

Common Point or Ground                                  Common Point or Ground

FIGURE 6-23 Two examples of using a “half wave” rectifier D1 to ensure that the LED does not experience any reverse bias voltage from the AC voltage source.

When using a peak hold capacitor, as shown in the “D1C1” circuit in Figure 6-23, be sure to increase the resistance value for R1 by about 3 times. The reason is that the average DC voltage at the positive terminal of C1 is higher than the half wave rectifier D1 circuit by about threefold. Thus, we see that the D1 C1 circuit’s LED current-limiting resistor R1″ = 3KΩ to 30KΩ, versus R1 = 1KΩ to 10KΩ in the circuit on the left side “D1 Rectifier.”

Alternatively, one can connect an LED or diode in “counter” parallel to the LED as shown in Figure 6-24 to safely limit the reverse LED voltage. Note that the protec- tion LED2 is connected with its cathode to the anode of LED1. In the D1 protection diode example, D1’s cathode is connected to the anode of LED1, and D1’s anode is connected to LED1’s cathode. In either example, the reverse bias voltage is limited to about –2 to –3 volts via LED2, and –0.7 volts via D1, which are both safe reverse bias voltages for LED1.

Two LEDs “Anti” Parallel                                       D1 Protection Diode

Tekstvak: 1Tekstvak: 2Tekstvak: 2Tekstvak: 1Red LED2


Tekstvak: 1Tekstvak: 2Silicon

Common Point or Ground                                Common Point or Ground

FIGURE 6-24 Two examples of protecting LED1 via LED2 and D1 wired in “anti- parallel.”

It should be noted that in Figure 6-24 LED2 lights up only during the negative cycle of the input AC voltage, and LED1 turns on only during the positive cycle of the AC voltage. This circuit can be used to identify the polarity of any incoming volt- age. For example, if the AC voltage source is replaced with a DC voltage, the two LEDs “anti-parallel” circuit will readily identify the polarity.

On the left-side circuit of Figure 6-24, LED2 can be any type of LED, such as IR (infrared), red, green, yellow, blue, or white.

And on the right-side circuit of Figure 6-24, D1 can generally be almost any diode including small signal, Schottky, or rectifier diode. If in doubt, just use a 1N4000 series rectifier (e.g., 1N4002).

Some Keys Points About Light Emitting Diodes

•     Light emitting diodes (LEDs) can be damaged by excessive forward bias current. Normally, the LEDs are driven at about 20 mA or less using a current-limiting resistor or some other circuit. See Chapter 7.

•     LEDs can also be damaged by reverse voltage by having a negative voltage from anode to cathode. If the negative voltage is limited to between 0 volts to

–4 volts, this will be OK. An “anti-parallel” connected diode or LED can be used to protect the LED in question.

•     Different color LEDs generally cannot be connected in parallel because their turn-on voltages are different, which will cause “current hogging” for the LED that turns on with the lowest voltage. For example, if blue and yellow LEDs are connected in parallel, almost all the current will go to the yellow LED. This is because a yellow LED has a turn-on voltage around 1.8 volts, whereas a blue LED requires a higher 2.4 volts.

•     If you need to have matched turn-on voltage LEDs, purchase them on a tape, as shown in Figure 6-21.

This concludes Chapter 6. In Chapter 7, we will explore transistors, where they can be used as current source and amplifying devices.

               CHAPTER 7

Bipolar Junction


In previous chapters, we have dealt with two terminal electronic components such as resistors, capacitors, diodes, and LEDs. We will now explore the commonly used bipolar junction transistors (BJTs, also known as bipolar transistors), which are three-terminal devices. They are in many do-it-yourself (DIY) projects and manu- factured products.

Bipolar Junction Transistors

The first transistor was invented at Bell Laboratories in 1948. It had a point contact type of construction that was difficult to manufacture. The point contact transistor technology was replaced by the (alloy) junction transistor that was more easily man- ufactured. Alloy junction germanium transistors such as the CK722, 2N107, 2N109, 2N408, 2SB32, 2SB54, 2SB56, 2SB75, 2SB77, 2SB113, 2SB115, AC125, AC126, OC71,

and OC72 were quite prevalent from the mid-1950s to the late 1960s.

So, what is a transistor? The bipolar transistor is a three-terminal device that allows controlling large amounts of output current by a small change in input cur- rent or voltage. Thus, a bipolar transistor can be used as an amplifier where a small input signal can provide a larger signal at its output, with the “help” of a power sup- ply. That is, transistors by themselves cannot amplify or control large electrical cur- rent unless there is a supply voltage included in the transistor circuit.

The three terminals to a bipolar transistor are:

•     The emitter as identified by the letter “E” or “e”

•     The base as identified by the letter “B” or “b”

•     The collector as identified by the letter “C” or “c”


See Figure 7-1 for bipolar transistors with different plastic packages. These range from small signal transistors (TO-92) to power transistors (TO-220 and TO-247).

FIGURE 7-1 Various transistors. Note the different pin outs for the two TO-92 transistors.

Most plastic small-signal transistors (e.g., < 500 mA collector current) with 2N or PN prefixes, such as 2N3904, 2N4124, 2N4401, PN2222 or 2N5089, have the pin out sequence E, B, C as shown in the first transistor on the left in Figure 7-1. This left-to-right emitter, base, collector sequence is taken as you look at the transistor with the flat side up.

In Europe, transistors that generally have a prefix of BC, such as BC548, BC556, or BC239C (as shown in Figure 7-1, second transistor from the left side), and other part numbers with a P2N prefix, such as P2N2222A and P2N2907A, have a “reversed” pin out sequence of C, B, E. The left-to-right collector, base, emitter sequence is taken when you look at the transistor flat side up.

For the two power transistors shown in Figure 7-1, the sequence, taken left to right with the part number side up or facing you, is base, collector, emitter, or B, C,

E. Note that the metal tab in the TO-220 transistor actually extends further down in the back of the transistor. This metal tab is connected to the middle lead or collector of the TO-220 power transistor. In general, with very few exceptions, it is very important to insulate or to ensure that the metal tab is not connected to ground or to some other point of a circuit. TO-220 transistors have maximum collector cur- rents of about 10 amps. However, for many power applications such as driving loud- speakers or motors, the TO-220 transistor should be mounted on a heat sink with a thermally conductive insulated sheet or washer.

The TO-247 transistor in Figure 7-1 can handle collector currents greater than or equal to 15 amps. Not shown but in its back side is an exposed metal piece that is also connected to the middle lead or collector of the power transistor. Again, if the transistor is delivering high power (> 5 watts) to a load such as a motor, loudspeaker, or lamp assembly, a heat sink is required. The TO-247 transistor would be mounted on a heat sink via a thermally conductive insulated sheet or washer. See Figure 7-2 for various TO-220 heat sinks.

FIGURE 7-2 Various TO-220 heat sinks with an example TO-220 voltage regulator mounted on the bottom-left side.

Transistors also come in metal packages, and the two most common ones are TO-18 and TO-5 (see Figure 7-3). These transistors have protruding tabs, which identify the emitter terminals.

FIGURE 7-3  TO-5 transistors on the left and the smaller TO-18 transistors on the right.

In metal case transistors the middle lead will usually be the base terminal. To find the emitter lead, find the tab that is located next to it. The remaining lead oppo- site the emitter terminal is the collector. Another way is to look at the bottom of the metal case transistor with the middle lead pointing to “12 o’clock.” The emitter lead will be to the left of “9 o’clock,” and the collector lead will be on the right that is analogous to a “3 o’clock” position.

Generally, metal case small and medium collector current transistors are not used as much, and they have been replaced with plastic package transistors. One reason is cost, and another is that they only offer a little more collector current capa- bility than some plastic case transistors of the similar sizes. What you will see, how- ever, is that metal case transistors were used often in older equipment (e.g., older Hewlett Packard solid state signal generators, vintage stereo amplifiers, tuners, and receivers, or ham radio transceivers, transmitters, and receivers). So, it is good to know about metal case transistors when you are fixing or modifying older circuits.

It is very important to know the pin out for each transistor you use because none of the three terminals is interchangeable. For example, if you reverse the collector and emitter leads, your circuit will not work properly. In other cases, mis-wired tran- sistor leads can result in damage.

In general, a damaged transistor usually has two characteristics:

1.            Shorted collector to emitter junction, which when measured with an ohm meter, the normally “open” or infinite ohms reading goes to a low resistance reading of typically < 10Ω.

2.            Open base to emitter junction, which means if you use your DVM’s diode voltage testing function, you do not get typically 0.6 volts to 0.7 volts, but instead get an “error” reading.

However, because most DVMs have a transistor tester, you can just test the tran- sistor. For power transistors, you can solder wires to their terminals for easier testing in a DVM.

If you have never seen a transistor in a circuit and want to just test it, what would you look for? The DVM tests for current gain known as Hfe, hFE, beta, or the Greek letter beta, β. Generally, a good transistor will have a current gain of 10 or more.

Current gain is defined as the ratio of the collector current, IC, to the base cur- rent, IB. That is:


5 hFE

5 beta 5 β 5 IC


Or put in another way, IC 5 β IB. Typically, 15 < β < 500.

For example, if we have 1 μA of base current, IB, and the β 5 250, then the col- lector current:

IC 5 250 3 1 μA


IC 5 250 μA

Typically, small signal (TO-92) and metal case transistors (e.g., TO-5 and TO-18) have ≥ 20 current gain. If the transistor is mis-wired and connected backwards in terms of the collector and emitter terminals, then you will see a substantial current gain decrease to typical numbers of 0 < β < 4.

See Figures 7-4 and 7-5, which show testing a 2N2906 (TO-18) transistor.

FIGURE 7-4 A TO-18 transistor with its emitter tab pointed up toward “E” with the DVM selector knob set to hFE. Here, we see the transistor has a current gain, hFE = 78.

There are two types of bipolar transistors, NPN and PNP. To identify the tran- sistor, you can plug the test transistor into the NPN socket and test, and then do the same in the PNP socket. Only one socket will give you a correct reading, which is usually a number between 15 and 1000. The other way to identify the transistor is to read the part number printed on the case and look up the data sheet. For exam- ple, the transistor being tested is a 2N2906, which its data sheet identifies as a PNP transistor.

Now, should you swap or reverse the collector and emitter leads, while having the base lead correct, you will find that the current gain is hFE 5 1. See Figure 7-5.

FIGURE 7-5 The same transistor tested “backwards” with the emitter inserted in the collector, C, portion of the socket and the transistor’s collector lead inserted into the E portion of the socket. Note that the tab that identifies the emitter is pointing down.

The resulting current gain is very low at hFE = 1 when the transistor is wired backwards this way.

What Happens When a Transistor Is Damaged

Transistors, just like diodes, share a couple of common specifications. One is the maximum voltage, and the other is the maximum current. When either of these specifications is exceeded, the transistor can be damaged. The typical blown or dam- aged transistor suffers from any combination of a short circuit or low resistance between the collector and emitter that can be verified with an ohm meter. Make sure to test the collector and emitter leads with the test leads one way and then the other— for example, the red test lead connected to the collector and black test connected to the emitter. Then switch the ohm meter’s leads where the black test lead is connected to the collector and the red test lead is connected to the emitter. Note the two resis-

tance readings, which should be <20Ω if the transistor is damaged via a short from collector to emitter.

A damaged transistor can have an open circuit between the base and emitter terminals. This can be verified by using the DVM’s diode voltage test as shown in Figure 7-6.

FIGURE 7-6 Testing the base-to-emitter voltage with the DVM where the one on the left is good with a forward base-to-emitter voltage reading of ~0.768 volt in three digits. But the one on the right is damaged with a single digit “1” that indicates an open circuit. If the reading is not in the 0.7-volt range, reverse the test leads just in case you are testing a transistor of the opposite polarity (e.g., NPN versus PNP, or PNP versus NPN). It is sometimes easy to incorrectly think of which type of transistor you are testing because

of a misread transistor part number.

Schematic Symbol of NPN and PNP Transistors

In most circuit schematic diagrams, transistors are designated with a “Q” prefix such as Q1, Q2, Q3, etc. (See Figure 7-7.) Alternatively, sometimes the transistors are labeled with a prefix of “TR,” which results in TR1, TR2, TR3, etc., or sometimes with a dash in between like TR-1, TR-2, TR-3, and so on.

FIGURE 7-7 Schematic symbols for NPN (Q1) and PNP (Q2) with their terminal’s emitter, base, and collector labeled equivalently as E, B, and C.

NOTE: The arrows in the emitter terminals show the direction of positive current flow.

Applying a DC Voltage to the Base of the Transistor to Provide a Constant Current Source

To reiterate, the transistor is an electrical current controlling device. In most cases, it provides current gain and/or voltage gain from an input to an output. Let’s take a look at several circuits where the transistor works as a constant current source. In the first examples, a small base current, IB, at the “input” results in a large collector cur- rent, IC, at the “output.” See Figure 7-8.

FIGURE 7-8 A constant current source via collector output current IC. A variable power supply voltage source, BT1, is applied directly to the base-emitter junction, which can damage the transistor Q1.

In Figure 7-8 a positive voltage, BT1, is applied to the base with respect to the emitter. And the output current, IC, requires a positive voltage to the collector with respect to the emitter.

A basic principle of biasing a transistor is shown in Figure 7-8, which is not nor- mally done in practice because the transistor can be easily damaged this way. Biasing is a way of turning on the transistor Q1 via applying a bias voltage, BT1, across the base and emitter terminals. In this case variable voltage source BT1 is turned up to a voltage generally in the range of 0.6 volts to 0.7 volts to generate a collector current IC at the output. However, in practice this method is rarely used because if BT1 exceeds 0.8 volts, the transistor may burn out and either the collector and emitter

terminals get shorted out due to excessive collector currents and/or the base-emitter junction becomes like an open circuit. This happens because the bipolar transistor has an exponential relationship between the base-emitter voltage and collector cur- rent. So, what does this really mean? It means that a small increase in base-emitter voltage (e.g., from BT1) will result in a large increase in collector current.

For example, if a transistor has 1 mA collector current for a BT1 set to 0.600 volt, the collector current will increase to 10 mA for BT1 5 0.660 volt, and 100 mA when BT1 5 0.720 volt. As you can see, every increase in 60 mV causes a change in collec- tor current by 10-fold.

Another reason not to directly drive the base-emitter junction directly with a voltage source is that every transistor has slightly different turn-on voltage. Generally, a power transistor will have slightly lower turn-on voltage. But even transistors of the same type (e.g., small-signal) or same part number can have base-emitter turn-on voltages that are different by 60 mV or more due to the transistor manufacturing process. This means that without sorting or measuring each transistor, you can get a 10:1 variation in collector current for the same base-emitter bias voltage.

Just like LEDs, transistors have maximum base current and collector current rat- ings. Generally, we look for the maximum collector current specification that is found in a data sheet.

For example, a very popular general purpose NPN transistor, 2N3904 (TO-92 plastic case, see Figure 7-1), has a maximum collector current rating of about 200 mA. For higher currents, a popular NPN power transistor is the MJE3055T transis- tor (TO-220 case, see Figure 7-1) that has a 10-amp collector current rating.

In practical transistor circuits, generally the base and/or emitter will include a resistor to help “stabilize” the collector current, which allows for plugging in different transistors while providing essentially a reliable or known collector current. Figure 7-9 shows simple series base resistor biasing. This type of circuit was used often in the early days of transistor circuits back in the 1950s. It can still be used today but there are problems with this circuit because β can vary from one transistor to another, which means the collector current will vary for a given fixed base current.

FIGURE 7-9 Biasing a transistor via a series base resistor, R1, to set a collector current IC.

To prevent accidental damage to the transistor due to excessive collector current, the series base resistor R1 isolates BT1 from connecting its voltage directly to the base-emitter terminals. The collector current IC then can be calculated in the fol- lowing manner.

Again, we need to reiterate that current gain is the ratio of IC (collector current) to IB (base current), or:

current gain 5 IC


There are two commonly used symbols for transistor current gain, β and hFE, which are one and the same. Thus:

β 5 hFE

5 IC


With a little algebra, we can find the collector current if we know the current gain and base current. In this book β and hFE are interchangeable. Thus:

IC 5 β IB

For example, if β 5 hFE 5 100 and IB 5 10 μA, then IC 5 100 3 10 μA or

IC 51000 μA 5 1 mA.

In Figure 7-9, the base current, IB, is the current flowing through R1. We can determine the current flowing through R1 by finding the potential difference across R1.

IB 5 (BT1 – VBE)/R1

If we know the battery voltage BT1, and approximate that VBE is in a narrow range of 0.6 volt to 0.8 volt, we can use a value in the middle such as 0.7 volt for VBE. Thus, IB 5 (BT1 – 0.7 volt)/R1.

IC 5 β IB so:

IC 5 β (BT1 – 0.7 volt)/R1

For example, suppose R1 5 100KΩ and β 5 100, with BT1 5 1.5 volts.

IC 5 100 (1.5 volts – 0.7 volt)/100KΩ 5 100 (0.8 volt)/ 100KΩ 5 80 volts/100KΩ

IC 5 0.0008 amp or 0.8 mA or 800 μA

If the β 5 300, then IC 5 300 (1.5 volts – 0.7 volt)/100KΩ 5 0.0024 amp or

2.4 mA.

The current gain β or hFE is usually specified by the transistor’s data or specifica- tion sheet. Also, β or hFE normally has a wide range such as 3:1. For example, the β can vary from 50 to 150 for some transistors, and 100 to 300 for others.

NOTE: In Chapter 6 we stated that an LED is generally not to be driven directly with a voltage source such as a power supply. This applies the same to any base-emitter junction. A base-emitter junction directly driven with a low-resistance power supply can result in damaging the transistor. Often, when excessive current has been applied to the base, the base-emitter junction becomes an open circuit and thus no longer conducts or allows base current to flow. This open circuit between the base and emitter, due to damage, will usually result in zero collector current.

We will now show examples of biasing a transistor via a series base resistor. See Figures 7-10 and 7-11, which illustrate driving LEDs.

FIGURE 7-10 A simple LED driver circuit using an NPN transistor current source.

The simple LED drive circuit in Figure 7-10 uses the existing collector supply voltage source, BT2, to drive the base resistor, R1. The base current with BT2 5 9 volts is then IB 5 (BT2 – 0.7 volt)/R1 and the LED current, which is the collector current, is then IC 5 ILED 5 βIB 5 β(BT2 – 0.7 volt)/R1.

For example, if β 5 50 and R1 5 50KΩ, then:

ILED 5 β(BT2 – 0.7 volt)/R1 5 [50 (9 – 0.7)/50,000] amp 5 [8.3/1000] amp 5

0.0083 amp


ILED 5 8.3 mA

The current gain in the simple circuit in Figure 7-10 can vary if you have the same part number transistors, such as 2N3904. Figure 7-11 shows an illustration where the β for Q1 5 129, and for Q2 β 5 239.

FIGURE 7-11 A simple LED drive circuit with a different current gain, b, for transistors Q1 and Q2.

Because of the two different current gains, β 5 129 and β 5 239 for transistors Q1 and Q2, the LED currents for LED1 and LED2, ILED1 and ILED2, are proportionally different as well.

ILED1 5 β(BT2 – 0.7 volt)/R1 5 [139(5.0 – 0.7 volt)/470,000] amp or

ILED1 5 0.00127 amp 5 1.27 mA

ILED2 5 β(BT2 – 0.7 volt)/R1 5 [239(5.0 – 0.7 volt)/470,000] amp or

ILED2 5 0.00218 amp 5 2.18 mA

The result of the current gain variation causes a change in LED brightness. See Figure 7-12.

FIGURE 7-12 The circuit built from Figure 7-11 shows that the LED on the left side via Q1where b = 139 is dimmer than the LED on the right side via Q2 where b = 239.

Improved Current Source Circuits

We can improve current source circuits to be less dependent on the current gain β by adding an emitter series resistor, RE, as shown Figure 7-13.

FIGURE 7-13 An improved circuit that is not b dependent to set the collector or LED current.

The collector current or LED current is just the emitter current IE 5 VE/RE because in a transistor the collector and emitter currents are essentially equal when used as a current source or amplifier circuit.


Resistors R2 and R3 form a voltage divider circuit that provide a base biasing voltage.

VB 5 BT2 [R2/(R2 + R3)]

To find VE, note that: VB 5 VBE 1 VE

VBE is the voltage across the base and emitter of Q1. As shown in Figure 7-13,

VBE 5 0.7 volt.

VB 5 0.7 volt 1 VE

Or put in another way in terms of VE:


VE 5 VB – 0.7 volt

We need to have VB greater than the turn-on voltage of Q1, which is 0.7 volt. But how much greater should we set it to? It depends on how much voltage we set for the emitter to ground voltage, VE. Typically, we want at least 0.5 volts so that any differ- ent lot of transistors with a 0.5-volt to 0.8-volt base emitter turn-on voltage will still work. But a rule of thumb would be something like VE ≥ 0.7 volt.

Because VB 5 0.7 volts 1 VE, and VE ≥ 0.7 volt:

VB 5 0.7 volt 1 ≥ 0.7 volt VB ≥ 1.4 volts

For example, suppose we want VB 5 1.7 volts with BT2 5 5 volts, which means VE 5 VB – 0.7 volt or VE 5 1.0 volts. Or in general:


Or put in another way:


For example, if we want 1 mA 5 0.001 amp 5 ILED1, and VE 5 1 volt, then:

RE 5 1 volt/0.001amp 5 1000Ω

Now let’s get back to solving for R2 and R3 when VE 5 1.0 volt.

VB 5 0.7 volt 1 VE

That is, if we measure the voltage across the base and emitter (VBE) and then measure the voltage (VE) at the emitter and ground, we will find VB 5 VBE 1 VE. VBE ~ 0.7 volt, so for 1-volt VE:

VB 5 0.7 volt 1 1.0 volt

VB 5 1.7 volts

Resistors R2 and R3 form a voltage divider circuit that provides a base biasing voltage with BT2 5.0 volts and RE 1000Ω. Again, in general:

VB 5 BT2 [R2/(R2 1 R3)]

However, via some algebra a more convenient equation can be derived as: [(BT2/VB) – 1] 5 R3/R2, and if BT2 5 5.0 volts and VB 5 1.7 volts, then we have:

[(5.0/1.7) – 1] 5 R3/R2

[2.94 – 1] 5 1.94 5 R3/R2


R3 5 1.94 3 R2

Now the question is, which resistors do we choose?

Can we make R2 5 1Ω and R3 5 1.94Ω? Yes, but you will be either burning up the resistors because the current through the resistors will be BT2/(R2 1 R3) 5 5 volts/2.94Ω 5 1.7 amps, or you will be wasting power like crazy.

Power 5 P 5 5 volts 3 1.7 amps 5 8.5 watts

A general rule of thumb would be to have R2 have the following range: RE ≤ R2 ≤ 10 RE

So, let’s take an example setting R2 5 10RE:

R2 5 10 3 1000Ω 5 10KΩ


R3 5 1.94 3 R2 5 1.94 3 10KΩ 5 19.4KΩ


R3 ~ 20KΩ

With the resistor values R2 5 10KΩ, R2 5 20KΩ, and RE 5 1000Ω, Figure 7-13’s current source will deliver a reasonably stable 1 mA given the battery voltage is 5.0 volts, even though the β of the transistor may vary 3:1 with a minimum β of 50. This is because the base current of Q1 no longer predominately determines the col-

lector current or LED current. Instead, it is the voltage VE (via VB) that sets the LED current for a given RE resistance value.

We can improve the current source further by “regulating” VB so that it is a con- stant voltage. See Figure 7-14 that uses a Zener diode as a reference voltage source, Vref.

FIGURE 7-14  A Zener diode current source circuit.

In this circuit, first determine the LED current, which is: VE 5 Vref – VBE


VBE 5 0.7 volt

VE 5 Vref – 0.7 volt And


ILED1 5 (Vref – 0.7 volt)/RE RE 5 (Vref – 0.7 volt)/ ILED1

As a general rule of thumb, we want the current through R3 to be ≥ 10% of ILED1. For example, we can set the current through resistor R3, IR3 5 (BT2 – Vref)/R3 5 10% ILED1, then:

R3 5 (BT2 – Vref)/(10% ILED1)

For example, suppose if we want ILED1 5 20 mA, and ZD1 5 1N4371 with a 2.7- volt Zener voltage Vref, then:

RE 5 (Vref – 0.7 volt)/ ILED1 5 (2.7 volts – 0.7 volt)/0.02 amp or RE 5 2.0 volts/0.02 amp 5 RE, which leads to:

RE 5 100Ω

BT2 5 5.0 volts, Vref 5 2.7 volts, ILED1 5 0.02 amp

R3 5 (BT2 – Vref)/(10% ILED1) 5 (5.0 volts – 2.7 volts)/(10% 0.02 amp)

R3 5 3.3 volts/0.002 amp 5 1650Ω ~ 1.6KΩ R3 ~ 1.6KΩ

In Figure 7-14, there is one precaution to take in terms of making sure the tran- sistor is not in saturation. To ensure that it is really working in the forward active region, we need to have the collector voltage ≥ base voltage, or where the collector voltage is ≥ Vref in this case.

The collector voltage in Figure 7-14 is VC 5 (BT2 – VLED). From Chapter 6, we can summarize the following:

•     For red, yellow, and most green LEDs, VLED ~ 1.8 volts to 2.0 volts (2.0 volts worse case)

•     For blue, white, and green phosphor LEDs, VLED ~ 2.7 volts to 3.0 volts (3.0 volts worse case)

•     If Vref 5 2.7 volts, then VC ≥ 2.7 volts for a red LED and BT2 5 5.0 volts. (We will use worse case numbers to be on the safe side.)

•     Thus, VC 5 (5.0 volts – 2.0 volts) 5 3 volts ≥ VB 5 Vref 5 2.7 volts.

So, we are “fine” using the circuit for red, yellow, and most standard green LEDs. However, if we use a white or blue LED, VC 5 (5.0 volts – 3.0 volts) 5 2.0 volts, which is less than Vref at 2.7 volts, then the transistor will most likely be in the satu- ration region. This means that excessive base currents forming a larger than expected voltage drop across R3 will cause Vref to drop and the desired LED current will be lower than expected. Essentially, when the transistor is in the saturation region by having VC < VB, the current gain is no longer β, but some number much less. So how do we fix the problem if we want to put in a white LED for LED1 while not saturating the transistor Q1? One way is by using a 1N4678 1.8-volt Zener diode, which results in Vref 5 1.8 volts and RE     56Ω and R3 can still be 1.6KΩ. The other way with Vref 5 2.7 volts is just to raise the battery or power supply voltage to 6 volts

or more.

However, in many cases we can replace the 1.8-volt Zener diode with three 1N914 small-signal diodes in series, or use a red LED as shown in Figure 7-15.

FIGURE 7-15 A practical circuit using a red light-emitting diode, LED2 in place of a Zener diode for Vref, a voltage reference source.

Almost any LED can be used as a voltage reference for biasing purposes. Just remember that standard infrared, red, yellow, and green (non-phosphor) LEDs have turn-on voltages of about 1.6 volts to about 2.0 volts. And the blue and white LEDs have about a 2.7-volt to 3.0-volt turn-on voltage. The LED turn-on voltage is then used to provide a reference voltage for biasing.

Figure 7-15 shows a red LED as a voltage reference source to supply about 1.7 volts 5 Vref. The emitter voltage, VRE 5 1.7 volts – 0.7 volt (VBE) ~ 1.0 volt across resistor RE. The voltage, VRE, across resistor RE then sets up the transistor’s emitter current. And essentially the emitter current is equal to the collector current, which in turn is the current flowing through LED1.

If RE 5 100Ω, then the LED current is 1.0 volt/100Ω 5 10 mA 5 IC 5 ILED1. Now if we work backwards to determine the base current, IB, that is equal to IC/β. And if β 5 50, IB 5 0.01amp/50 5 0.2 mA 5 IB. With BT2 = 5 volts, the current flowing through R3 is [5.0 volts – VLED2] divided by 1600Ω. With VLED2 5 1.7 volts, we have 3.3 volts/1600Ω 5 2.06 mA 5 current flowing through R3. We can think of the current flowing through R3 branches out to two DC currents:

•     A current flowing into LED2, ILED2

•     And a current flowing into the base of transistor Q1, IB

We know that the base current, IB 5 0.2 mA 5 IC/β 5 0.01amp/50, when β 5 50.

Of that 2.06 mA flowing through R3, 0.2 mA goes to the base of Q1 via IB, while the remaining current, ILED2, flows into LED2. To find ILED2, we know that:

3.3 volts/1600Ω 5 2.06 mA 5 current flowing through R3 5 IB 1 ILED2

2.06 mA 5 IB 1 ILED2 But IB 5 0.2 mA, so:

2.06 mA 5 0.2 mA 1 ILED2

By algebra, subtracting 0.2 mA from both sides of the equation:

2.06 mA – 0.2 mA 5 ILED2

ILED2 5 1.86 mA

The main thing to take away from this is that there is sufficient current to turn on LED2. Generally, any LED current between 0.5 mA and about 20 mA will work fine.

What Happens When Things Go Wrong

Probably the most common mistakes in LED, diode, or transistor circuits, are reversed component connections. We will look at some examples and see what this will do to other parts of the circuits, starting with Figure 7-16.

FIGURE 7-16 Voltage reference LED2 is reversed and can cause LED1 to burn out.

With LED2 connected in “reverse bias” mode, it will not light up and it is essen- tially an open circuit that causes VLED2″ >> 1.7 volts. This means that BT2’s voltage via resistor R3 will supply Q1’s base voltage, VB, which will be much higher than

1.7 volts. With VB >> 1.7 volts, the LED1 current ILED1 5 (VB – 0.7 v)/RE and LED1 will have excessive current flowing through it. For example, if RE 5 75Ω and VB is

supposed to be 1.7 volts but someone inadvertently reversed the leads and VB        3.3 volts via R3, then:

ILED1 5 (VB – 0.7 v)/RE 5 (3.3 v – 0.7 v)/75Ω 5 34 mA, for the reversed LED2 connection

The actual voltage VB when LED1 is reverse biased is difficult to calculate off- hand because we often do not know the actual β of Q1. Also, the voltage of BT2 enters into this calculation. As a troubleshooter, we do not have to “bother” with trying to figure out the circuit analysis for an incorrectly wired circuit. We just need to solve the problem and restore the circuit back to proper operation. That is, don’t waste too much time analyzing circuits with mistakes. We have to get it working and move on.

To fix this, you connect LED2 properly (LED2 anode to base Q1, and LED2 cathode to ground; the LED’s long lead 5 anode) as shown in Figure 7-15, and then VB ~ 1.7 volts so that:

ILED1 5 (VB – 0.7 v)/RE 5 (1.7 v – 0.7 v)/75Ω 5 13.3 mA

Now let’s look at Figure 7-17 with Q1’s collector and emitter leads reversed.

FIGURE 7-17 With Q1’s emitter and collector leads reversed, the current gain is very low with b ≤ 4.

With the collector and emitter leads reversed, transistor Q1 has low β and starts pulling extra current via resistor R3. This causes a large voltage drop across R3, which is IB 3 R3. For example, if RE 100Ω and β 52 (due to having the emitter- collector terminals reversed), then what is VB? It has to be lower than 1.7 volts,

which will mean that LED2 does not turn on due to the low β 5 2. We can approxi- mate VB as follows: If the desired voltage, VB 5 1.7 volts, then VRE 5 1.7 volts – 0.7 volt 5 1 volt, then the desired emitter current is 1 volt/100Ω 5 10 mA.

NOTE: For small values of b, generally b < 10, we will need to use the more exact equations relating IB and IC and IE. They are: IE = (b + 1)IB, IB = IE/(b + 1), and we still have IB = IC/b , which leads to: IC = IE [b/(b + 1)]. For example, if b = 9, IC = IE [9/(9 + 1)] = IE(0.90).

If IE 5 10 mA and β 5 3, then the base current:

IB 510 mA/(β 11) 5 10 mA/(3 1 1) 5 10 mA/(4)

IB 5 2.5 mA

Thus, the voltage drop across R3 5 1600Ω 3 2.5 mA 5 4.0 volts. VB 5 BT2 –

(voltage drop across R3). VB 5 5 volts – (4.0 volts) 5 1 volt. At 1 volt 5 VB, this means that LED2 does not turn on due to LED2’s turn-on voltage being 1.7 volts. This leaves VRE 5 VB – 0.7 volt 5 0.3 volt.

The emitter current IE 5 0.3 volt/100Ω 5 3 mA.

The “exact” collector current IC 5 IE [β/(β 1 1)] 5 3 mA [3/(3 1 1)]

IC 5 3 mA [3/4] 5 3 mA (0.75).

IC 5 ILED1 5 3 mA (0.75) 5 2.25 mA, which is smaller than the 10 mA of LED1 current if the transistor Q1 was correctly connected.

There are two things to take away from connecting the transistor incorrectly in Figure 7-17.

•     The reference voltage light-emitting diode LED2 may go dim or turn off.

•     The LED1, which is expected to be driven with 10 mA, will be driven with a lower current and will give out much less light.

There are times when we assume that the transistor’s pin out has an order of emitter, base, and collector (EBC). However, there are some transistors such as PN2222 (EBC) that has a similarly named part number P2N2222 (CBE) that has reversed emitter and collector leads. Generally, in the United States common TO-92 plastic case transistors such as 2N3904, 2N3906, 2N4124, and 2N4126 have emitter, base, collector pin outs (EBC). Whereas often used TO-92 transistors in Europe such as BC213, BC238, BC308, BC547, BC557 have pin outs of collector, base, emitter (CBE).

Now let’s look at another example where a Zener reference diode is reversed (see Figure 7-18).

FIGURE 7-18  Zener diode ZD1 is reverse connected, which results in a dimmed LED1.

With a Zener diode connected in reverse, it gives a voltage, 0.7 volt, similar to a forward biased diode. Because of this, VB ~ 0.7 volt, and VRE 5 VB – VBE, with VBE 5 0.7 volt. So VRE 5 VB – 0.7 volt 5 0.7 volt – 0.7 volt ~ 0 volt 5 VRE. As a result, LED1 is either turned off or being driven with a very small current.

What happens if we reverse LED1’s terminals? We know that it will not light up.

But what happens to the rest of the circuit? See Figure 7-19.

FIGURE 7-19 LED1’s anode and cathode leads are reversed, which leads to some very interesting analysis for VB and VE.

We see that with LED1 wired incorrectly, it will not light up. This means it’s essentially an open circuit, which disconnects Q1’s collector lead to any DC path to BT2. See Figure 7-20.

FIGURE 7-20  An equivalent circuit to Figure 7-19 using a diode “model” in place of Q1.

To determine VB, the base-emitter junction of Q1 now acts like a diode. There is no such thing as β or current gain because the collector lead is disconnected. If we make an estimate that VLED2 is less than its turn-on voltage (1.7 volts), then we can safely say that there is no current flowing into LED2. This means the current is flow- ing from BT2 via resistor R3, diode “Q1”, and RE to ground; and no current flows into LED2.

One way to calculate VB is by finding the current IB flowing though RE. The current, IB, flows into diode “Q1” and it also flows through RE.

Let’s work out a variation of a loop equation. That is the voltage:

BT2 5 Voltage across R3 1 diode voltage of “Q1” 1 voltage across RE. The diode voltage of Q1 can be approximated as 0.7 volt.

If you use a volt meter and try measuring BT2 first and then measure the volt- ages across R3, “Q1” diode, and RE, you will find that summing up all the voltages across R3, “Q1” diode, and RE will equal the voltage of BT2 5 5 volts.

With the following:

•     Voltage across R3 5 IB 3 R3

•     Voltage across RE 5 IB 3 RE

•     Diode voltage of “Q1” 5 0.7 volt (see Figure 7-20)

we can now solve for IB, and once that is done we can find IB 3 RE 5 VRE, which will lead to VB 5 0.7 volt 1 VRE.

Let BT2 5 5.0 volts, R3 5 1600Ω, and RE 5 100Ω.

BT2 5 IB 3 R3 1 0.7 volt 1 IB 3 RE

5.0 volts 5 IB 3 1600Ω 1 0.7 volt 1 IB 3 100Ω

5.0 volts – 0.7 volt 5 IB 1600Ω 1 IB 3 100Ω 5 IB (1600Ω 1 100Ω) 5

IB 3 (1700Ω) 5 5.0 volts – 0.7 volt 5 4.3 volts

4.3 volts 5 IB (1700Ω) or by dividing by 1700Ω on both sides of the equation: (4.3 volts/1700Ω) 5 2.53 mA 5 IB

IB 3 RE 5 VRE 5 2.53 mA 3 100Ω 5 0.253 volt 5 VRE

VB 5 0.7 volt 1 VRE 5 0.7 volt 1 0.253 volt

VB 5 0.953 volt

And as confirmed, VB is less that the 1.7 volts that is needed to turn on LED2.

Insufficient “Headroom Voltage” for the Transistor

We now need to discuss the limitations of transistor current sources. This means we need to pay attention to the collector-base voltage. See Figure 7-21.

FIGURE 7-21 An LED drive circuit on the “edge of disaster” due to VC ~ VB. If BT2’s voltage drops slightly (e.g., by ~ 0.5 volt), the circuit will start to fail. For proper operation, it is preferable that the collector voltage VC is greater than the base voltage VB, or in other words VC > VB.

In Figure 7-21, as shown, the LED2 voltage is about 2.0 volts 5 VB, and we replaced LED1 with a blue or white LED, which has a turn-on voltage of 3.0 volts. This means the collector voltage:

VC 5 BT2 – VLED1 5 5.0 volts – 3.0 volts

VC 5 2.0 volts

Generally, to play it safe, VC ≥ VB. In this case, the LED circuit is still workable. However, if the battery voltage, BT2, drops from 5.0 volts to 4.5 volts, then VC 5 4.5 volts – 3.0 volts 5 1.5 volts. This means VC < VB, or in other words, VB > VC. When VB > VC, the transistor Q1 is either approaching the saturation region, or is already in the saturation region. For current source and amplifying circuits, we want to avoid transistors with these conditions because the β will start dropping.

So, one fix to Figure 7-21 is to lower VB while maintaining a “reasonable voltage” at VE (e.g., VE ≥ 0.5 volt). See Figure 7-22.

FIGURE 7-22 Base voltage VB is lowered by replacing the green LED 2-volt reference voltage with a lower one at 1.4 volts via two series connected diodes, CR1 and CR2.

With this modification, VB 5 1.4 volts, while VC 5 2.0 volts. Note that the 1.4 volts developed by these two diodes stay relatively constant should BT2 drop in volt- age to 4.5 volts.

Hence, even if BT2 drops to 4.5 volts, VC 5 1.5 volts > VB 5 1.4 volts, which provides a little safety margin.

Alternatively, the two diodes may be replaced with a 620Ω resistor, R4, to pro- vide 1.4 volts for BT2 5 5.0 volts. However, VB will drop proportionally with BT2. For example:

BT2 5 5.0 volts   VB 5 1.4 volts

BT2 5 90% 5.0 volts 5 4.5 volts  VB 5 90% 1.4 volts 5 1.26 volts

This is because VB ~ BT2[R4/(R3 1 R4)], neglecting base current IB, which is reasonable for a β ≥ 50.

Another option to avoid transistor Q1 from going into the saturation region is to simply increase the supply voltage BT2, if possible. For example, in Figure 7-21, where BT2 5 5.0 volt and VB 5 VC 5 2.0 volts, having BT2  6.0 volts will result in VC 5 6.0 volts – 3.0 volts 5 3.0 volts, which is greater than the 2.0 volts of VB.

Sometimes Even a Correct Circuit Goes Bad

So far, we looked at only DC conditions, which is very important in troubleshooting circuits. However, even when the DC bias points should work, there can be other problems. This transistor LED circuit discussed so far acts as both current source circuit, but also as amplifier. See Figure 7-23, which can self-oscillate at a very high frequency (e.g., > 100 MHz).

FIGURE 7-23 A current source circuit that can oscillate at very high frequencies due to a long base lead.

Usually, this circuit will work fine. But if you replace the transistor Q1 with a much higher frequency device, such as a VHF (very high frequency) transistor (e.g., MPSH10 whose ft > 600 MHz), the circuit may oscillate at a very high frequency that causes a variation in LED1’s drive current when you touch the base lead.

A long wire connected to Q1’s base behaves like an inductor, which forms a reso- nant circuit with stray and parasitic capacitances included in the transistor. This can inadvertently form an oscillator circuit. To stop the “parasitic” oscillation, we have to make a lossy inductor by adding a series resistor. See Figure 7-24.

FIGURE 7-24 A series base resistor, Rbase is added in series to the terminal of Q1 to prevent parasitic oscillations.

Rbase usually has a value of 22Ω to 220Ω and should have very short leads itself. And Rbase should be soldered or placed very close to Q1’s base lead. The objective is to prevent the base lead of Q1 from being connected to a long wire.

Figure 7-24 also shows decoupling capacitors to ground, C1 and C2. Generally, you put a decoupling capacitor at your circuit that is receiving the power supply volt- age. C2 is added at the circuit near LED1, and not at BT2, which can be connected via long wires that behave like inductors. Decoupling capacitor C1 is placed across the voltage divider circuit R4 and R3 to reduce any high-frequency noise. Generally, most decoupling capacitors have values between 0.01 μf to 2.2 μf.


We have presented many circuits showing how the transistor works in a current source circuit. Also, several troubleshooting techniques have been illustrated. The gist of troubleshooting is to learn by experience more so than try to be an engineer and analyze everything. Basically, troubleshooting is detective work. You will need to look up data sheets to find the pins outs, but also other specifications like maximum current or voltage that the device can take before breakdown.

Chapter 8 will continue with simple transistor amplifiers based on the LED drive circuits.

               CHAPTER 8


Discrete Circuits

(Simple Transistor Amplifiers)

For this chapter we will be introducing transistor amplifiers that are “derived” from the LED circuits from Chapter 7. We shall be showing that these amplifiers have limitations such as amplitude gain (ratio of output signal to input signal amplitudes), waveform distortion (e.g., a sine wave input signal with a distorted waveform at the output), output swing, and output current. In general, an amplifier is a circuit that provides a larger voltage signal or larger current signal at the output when compared to the input signal. All amplifier circuits require some type of supply voltage such as a 9-volt battery or a 12-volt power supply.

Important Practical Transistor Specifications

It should be noted that amplifiers are imperfect in that they cannot supply infinite current nor infinite voltage at their outputs. The transistors themselves have limita- tions on maximum collector-to-emitter voltage and maximum collector current, along with maximum power dissipation. For example, the ubiquitous 2N3904 NPN transistor has a typical maximum collector-to-emitter voltage (VCE) of 40 volts, a 200 mA maximum collector current (IC), and a 200 mW power dissipation rating. Power dissipation is usually given by Pd 5 VCE 3 IC. For example in Figure 8-1, suppose the collector current driving the light-emitting diode LED1 is 20 mA IC, and we have 9 volts as the supply BT2 with a red LED1 with a 2-volt forward turn-on voltage 5 VF. BT2 5 9 volts and VF 5 2 volts so BT2 5 VF 1 VCE, and VCE 5 BT2 – VF or

VCE 5 9 volts – 2 volts, so:

VCE 7 volts

Pd 5 VCE 3 IC, which is the power dissipation of transistor Q1


Pd 5 7 volts 3 20 mA

Pd 5 140 mW, which is fortunately below the 200 mW power dissipation rating of the 2N3904

Although this example will work, as a rule of thumb, you want the power dissipa- tion to be less than 50 percent of the maximum rating. In this example, the transistor is dissipating at 70 percent of maximum power dissipation (via 140 mW/200 mW 5 70%), so we may want to use a higher wattage transistor, such as a 2N4401where Pd 5 500 mW maximum.

FIGURE 8-1  An LED drive circuit used for determining transistor Q1’s power dissipation.

Note that both VCE 5 7 volts and IC 5 20 mA are well within the maximum specifications of 40 volts and 200 mA, respectively, for the 2N3904 transistor. However, this example shows that we have to keep track of power dissipation even if the voltages and currents are well within specifications.

Simple Transistor Amplifier Circuits

The basic LED constant current source drive circuit shown in Figure 8-2 can be reconfigured to operate as an amplifier as shown in Figure 8-3.

NOTE: The base driving resistor is renamed as RB.

In Figure 8-2 the transistor’s base-to-emitter voltage is held constant via the base current driving resistor RB that is connected to the power supply. With this constant base-to-emitter voltage, a constant collector current is provided to the LED. However,

if we somehow are able to vary the base-to-emitter voltage by just a little, such as a variation of a few millivolts (mV), then the collector current will vary as well. See Figure 8-3, which is essentially the same circuit with a couple of modifications.

FIGURE 8-2  Constant current source to the LED via Q1’s collector current.

FIGURE 8-3  The constant current source LED drive circuit modified into an amplifier.

For this type of amplifier, the signals must be low level like the types of signals from microphones that produce < 10 mV to avoid distorted output signals. If larger signals are used such as an audio signal from your digital or CD player, then the input should be attenuated with a voltage divider circuit. Typical attenuation would lower the amplitude by 10 to 100 fold.

Although this circuit is “easy” to analyze, it has limited purposes on its own as an amplifier. However, some circuits do not require low distortion. This circuit can be used for a fuzz pedal distortion amplifier used in an electric guitar, or be used in an oscillator circuit, or be used as a mixer that deliberately generates distortion prod- ucts, such as in an RF mixer.

First DC Analysis: Capacitors = Batteries with Self Adjusting Voltages

Capacitors C1 and C2 can be viewed as “fast charge” batteries that charge to a voltage such that no DC current flows through them. Let’s take a closer look at the one- transistor amplifier with some voltages labeled. See Figure 8-4 first.

FIGURE 8-4 Capacitors C1 and C2 are charged to voltages “VBE” and “VCE” as marked.

When we look at Figure 8-5, the DC voltage at Vin is equal to the series voltages of BT_C1 and the voltage at the base-emitter junction of Q1. Because the polarities of BT_C1 and VBE are the same with equal voltages via the 1 side of BT_C1 con- nected to the 1 side of VBE, the voltage at Vin is 0 volts DC. This amounts to two back-to-back equal voltage sources connected in series such that the net voltage is zero. Here is another way to look at this. Suppose you have a two-cell flashlight. Instead of installing the 1.5-volt batteries correctly (in series) to provide 3 volts, you install them back to back, which gives zero volts instead.

FIGURE 8-5 Batteries BT_C1 and BT_C2 can be thought of as replacing C1 and C2 as voltage sources.

In the amplifier, if the VBE of Q1 5 0.70 volts, capacitor C1 charges up to exactly

0.70 volts in Figure 8-4. In Figure 8-5, we see if BT_C1 is 0.70 volt as labeled, then the DC voltage at Vin has to be zero volts due to series connection of back-to-back VBE voltages from the Q1 and C1.

Likewise, if C2 in Figure 8-4 is charged up to a voltage of VCE, we see in Figure 8-5 that the DC voltage at Vout has to be zero volts. The reason is that BT_C2 has an equal and opposing voltage to VCE, the voltage at the collector of Q1. Thus, the DC voltage at Vout has to be zero volts due to the back-to-back series connection of VCE from the transistor and VCE in C2 or BT_C2.

The capacitors C1 and C2 play two important roles. First, they block out DC cur- rents from Vin and Vout. Second, they pass through an AC signal such as an audio signal.

Second DC Analysis: Take Out the Capacitors to Find the DC Currents and DC Voltages

With the capacitors removed we can now more clearly see for DC analysis. Not only the capacitors C1 and C2 have been removed but the adjoining components R1 and R2 are now also no longer part of the circuit. What we are left with is just three com- ponents, RL, R2, and Q1, plus the power supply, BT2.

There are two reasons for determining the DC base and collector currents, IB and IC. Finding the expected base and collector currents will allow us to determine the collector-to-emitter voltage, which has to be > 0.7 volts in most cases. For exam- ple, if the collector current is too high or the voltage drop across RL is too high, then the transistor is in the saturation region, and the circuit will not amplify.

FIGURE 8-6 To calculate the DC currents and voltages, remove all capacitors that are C1 and C2.

To determine the collector current IC, first, we calculate for the base current IB. IB 5 (BT2 – VBE)/RB

For most good approximations, VBE 5 0.7 volt.

IB 5 (BT2 – 0.7 v)/RB

IB 5 (BT2 – 0.7 v)


Let’s take some examples pertaining to Figures 8-4 or 8-6:

If BT2 5 9 volts and RB 5 56kΩ, then IB 5 (9 v – 0.7 v)/56kΩ 5 (8.3 v)/56kΩ

IB 5 0.148 mA or IB 5 148 μA


For this example, we use a lower current gain transistor such as a 2N3903 where β 5 50. Since we already calculated the base current as IB 5 0.148 mA then:

IC β IB 5 50 3 0.148 mA

IC 5 7.4 mA

If we have RL 5 1000Ω, then we can now find VC, Q1’s collector voltage refer- ence to ground.

VC 5 BT2 – IC (RL)

VC 5 9 v – 7.4 mA (1000Ω) 5 9 v – 7.4 v

VC 5 1.6 volts

What happens if β   100?

Then IC β IB 5 100 3 0.148 mA 5 14.8 mA.

We find that VC 5 9 v – 14.8 mA (1000Ω) 5 9 v – 14.8 v 5 –5.8 volts 5 VC???. This is not possible since we only have a positive power supply and thus no negative volts can be generated. Instead, VC ~ 0 volts, and usually VC ~ 0.2 volt. That is, if the calculated value for VC is < 0.5 volt, usually the amplifier will not work because the transistor is now a switch that “shorts” the collector terminal to ground. What we have shown in this example is that if the current gain is too high, the transistor amplifier in Figure 8-4 can cause the transistor to turn into a switch. That is why a circuit like Figure 8-4 requires selecting the β of transistors in a narrow range, such as 30 < β < 50, or “tweaking” the resistance value of the base driving

resistor, RB.

For example, if we want to use a transistor whose β 5 100, then we should approximately double the value of RB from 56kΩ to 110kΩ with BT2 5 9 v and RL 5 1000Ω. This leads to:

IB 5 (BT2 – 0.7 v)/RB 5 (9 v – 0.7 v)/110kΩ 5 8.3 v/110kΩ 5 75.455 μA

IC 5 100 3 75.455 μA 5 7.545 mA

VC 5 BT2 – IC RL 5 9 v – 7.545 mA 3 1000Ω 5 9 v – 7.545 v 5 VC

VC ~ 1.45 volts, which is pretty close to VC 5 1.6 volts for RB 5 56kΩ and β 5 50.

It would be safer to aim for a slightly higher VC. We can have RB 5 120kΩ with β 5 100 and RL 5 1000Ω, which results in IB 5 69.167 μA and IC 5 100 3 69.167 μA

or IC 5 6.9167 mA.

VC 5 9 v – 6.9167 mA 3 1000Ω 5 9 v – 6.9167 v 5 VC

VC 5 2.08 v

In the real world, the specific transistor part number (e.g., 2N3904, 2N2222, etc.) will not only have a range of β, but the current gain β will change with temperature. So, the DC collector current and DC collector voltage will vary. In general, the tran- sistor amplifier in Figure 8-4 is more of a hobbyist or DIY (do-it-yourself) amplifier where you have to individually tweak it to a desired collector current. We will find in the next section that the DC collector current IC also determines the AC signal’s gain at Vout, and also the input loading resistance at Vin.

For now, we have:

VC 5 BT2 – IC (RL)


IC 5 β(BT2 – 0.7 v)

A more general formula would be:

VC 5 BT2 – β(BT2 – 0.7 v)(RL)


For a current gain β 50, BT2 5 9 volts, RB 5 56kΩ or 56,000Ω, and RL 5 1000Ω.

VC 5 9 v – 50 (9 v – 0.7 v)(1000) 5 9 v – 7.4 v 5 VC


VC 5 1.6 volts

And note the collector current:


IC 5 β(BT2 – 0.7 v) 5 50 3 0.148 μA 5 7.4 mA 5 IC

Alternatively, instead of using the general formula for VC, sometimes it’s more logical to find the base current, IB, first and then calculate the collector current, IC to find VC, the collector voltage.

If BT2 5 5 volts and RB 5 100kΩ, then IB 5 (5 v – 0.7 v)/100kΩ 5 (4.3 v)/100kΩ.

IB 5 0.043 mA or IB 5 43 μA

If β 5 78, then IC 5 β IB 5 78 (43 μA).

IC 5 3.225 mA

Let RL 5 820Ω

VC BT2  IC (RL) 5 5 volts – 3.225 mA (820Ω) 5 5 volts – 2.6445 volts 5 VC

VC 5 2.3555 volts

Finding the AC Signal Gain

The DC collector current is the most important DC characteristic to find for calcu- lating the AC signal gain. Via the DC collector current we can also calculate the input resistance.

Input resistance to an amplifier is important since the input signal source itself usually has an output source resistance or an optimum input load resistance. For example, an antenna may have a 50Ω source resistance. To achieve maximum power transfer, the input resistance of the amplifier should be 50Ω as well.

In another example, if your signal source is a dynamic microphone, then the input resistance should be typically 1000Ω or more. If you should connect the dynamic microphone to an amplifier with a 50Ω input resistance, then you will lose signal amplitude from the microphone.

Thus, we have to keep in mind the amplifier’s input resistance based on the application (e.g., RF amplifier, microphone preamplifier, sensor amplifier, etc.). This is particularly important when we have one amplifier’s output connected to the input of a second amplifier.

We can find the AC signal gain of this amplifier by noting the following:

•     Capacitors with “sufficiently large” capacitances to be like batteries are AC short circuits or zero ohm resistors for AC signals.

•     Power supplies are also treated as AC short circuits or zero ohm (0Ω) resistors with respect to AC signals.

At first glance, the second item may seem absurd. But it you think about it, if you probe for AC signals in a clean power supply or battery, you will find no AC signal. By definition, power supplies and batteries are DC voltage sources and thus cannot include any AC signal.

Likewise, if a capacitor has a large enough capacitance that it acts like a “battery” (e.g., see Figure 8-5 again), then again by definition a battery cannot include an AC signal; it only produces a DC voltage. Let’s take a look again at the “original” sche- matic in Figure 8-7.

FIGURE 8-7 The one-transistor amplifier from Figure 8-5.

Now let’s take a look at the AC analysis model shown in Figure 8-8.

FIGURE 8-8 The capacitors C1, C2, and battery (DC power supply) BT2 are “modeled” as AC short circuits or AC zero ohm wires.

Figure 8-8 can be further redrawn to look simpler or more intuitively familiar.

See Figure 8-9.

FIGURE 8-9 A redrawn version of Figure 8-8 to show a more intuitive idea of the one- transistor amplifier’s AC analysis circuit.

As you can see in Figure 8-9, the circuit looks a little “funny” in that there is no power supply. However, the schematic gives us an idea of how the AC signal at Vin will be amplified. Note that there is a resistance into Q1’s base, which is referenced to ground.

Finally, we can model Figure 8-9 in terms of a block diagram with the resistors R1, RB, RL, and R2. See Figure 8-10. Note that R_in base is now “modeled” as a resis- tor to ground along with R1 and RB.

FIGURE 8-10 A block diagram of Figure 8-9’s AC analysis circuit.

In order to complete the AC analysis circuit, we have to find two “unknowns,” resistor, R_in base, and the gain, “G.” See below for the formulas given that IC 5 DC collector current.

R_in base 5 β (0.026 v/IC)

G 5 –(IC/0.026 v) 3 RL || R2

RL 1 R2 
R1 1 RB 

Where RL || R2 5 RL 3 R2 5 Ro and R_par 5 R1 || RB 5 R1 3 RB

Sometimes it is convenient to express the term (0.026 v/IC) 5 re, where a handy starting point of IC 5 1 mA results in:

(0.026 v/0.001A) 5 26Ω 5 re

From this 1 mA starting point we can find rfor any other collector by a scaling factor such that ris inversely proportional to collector current. That is, for a given current:

5 0.001 A 3 26Ω

e                    IC

Since we often set collector currents in milliamps, mA, you can express the for-

mula this way where the collector current IC is in milliamps.

5 0.001 A 3 26Ω

e        IC mA

For example, if IC 5 2 mA, then 5 1 mA 3 26Ω 5 13Ω.

e          mA

And if IC 5 0.1 mA 5 100 μA, then 5  1 mA  3 26Ω 5 260Ω.

e          0.1 mA

For the input resistance R_in base: R_in base 5 β (0.026 v/IC) 5 β re

NOTE: R_in base is inversely proportional to IC. For example, if you decrease IC by tenfold, then R_in base increases by tenfold.

For example, if β 5100 and IC 5 1 mA, then 5 1 mA 3 26Ω.

e           1 mA

r5 26 Ω

R_in base 5 β r5 100 3 26Ω 5 2.6KΩ

If β 5100 and we have IC 5 0.1 mA, then R_in base 5 β r5 26KΩ. For the gain factor G:

G 5 –(IC/0.026 v) 3 RL || R2

NOTE: G is proportional to the DC collector current IC. So, if you increase IC by tenfold (e.g., 0.1 mA to 1 mA), the gain, G, goes up by tenfold. And if you decrease IC by tenfold, G, decreases by tenfold.

In this amplifier given current gain β, if you increase the DC collector current the input resistance to the base, R_in base, decreases while the gain, G, increases.

A decreasing R_in base can cause the input signal to be loaded down and thus attenuated when we increase the DC collector current IC. One way to offset this is to pick a transistor that has a higher β. For example, a 2N3904 may have a typical β of 150, but a 2N5088 has higher typical β of 300. So, if you want to increase the collector current twofold but keep the same R_in base resistance, you can select a transistor with twice the β. To keep the same DC collector current IC, this also means that RB would be increased about twofold in resistance because of the higher current gain (e.g., twice the β).

Now let’s take one of the previous examples concerning DC analysis:

BT2 5 5 volts, RB 5 100kΩ, then IB 5 (5 v – 0.7 v)/100kΩ 5 (4.3 v)/100kΩ

IB 5 0.043 mA or IB 5 43 μA

If β 5 78, then IC 5 β IB 5 78 (43 μA) 5 3.225 mA.

IC 5 3.225 mA 5 0.003225A

RL 5 820Ω

R1 5 R2 5 220 KΩ

Let C1 5 C2 5 33 μf (with C1and C2 rated at ≥ 16 volts), RL 5 820Ω, R2 5 100KΩ, with IC 5 0.003225 A.

Gain Calculation

G 5 –(IC/0.026 v) 3 RL || R2

820 1 100K 

G 5 –(0.003225A/0.026 v) 3 820Ω ||100KΩ 5 –(0.124A/v) 3 820 3 100K Ω

820 1 100K 

820Ω ||100KΩ 5 820 3 100K Ω 5 813Ω

NOTE: The units A/v or amps per volt has a unit of (1/resistance) or 1/Ω.


G 5 –(0.124 A/v) 3 813Ω 5 –100.85 5 Vout


Vout 5 G 5 –100.8

The minus sign indicates an inverted AC signal, or a signal that is 180 degrees out of phase.

For example, if Vin is a 1 mV peak sinewave, then Vout provides an inverted sinewave at 100.8 mV peak.

Base Resistance, R_in Base Calculation

R_in base 5 β (0.026 v/IC) 5 78 (0.026 v/0.003225A) 5 78 (8.06Ω)

NOTE that the units v/A or volts per amp has a unit of resistance or Ω.

R_in base 5 628.8Ω

Limitations of This One-Transistor Amplifier

We will now look at the one transistor amplifier in terms of input amplitude range. See Figure 8-11.

FIGURE 8-11 The one-transistor amplifier that requires a low-level input signal to avoid distortion.

Limited Input Amplitude Range

When Vin is driven with a low-impedance generator such as one that has a 50Ω source resistance, which is common in many function generators, this amplifier will start to distort with signals starting at about 1-mV peak sine wave. For example, a 10-mV peak signal that is 20 mV peak to peak at the input will result in Vout having a generally larger amplitude but also a waveform that is distorted with about 10 per- cent second order harmonic distortion. As an example, if the frequency is 1000Hz, the output signal will provide a signal at 1000Hz, and another one at 2000Hz at 10 percent of the amplitude at 1000Hz.

Because this amplifier can produce so much distortion with signal sources from portable music players or smartphones (> 100 mV peak), you would need to place a voltage divider circuit such as a volume control to adjust the level into Vin, the input terminal.

See Figure 8-12 for an amplifier with the following:

R1 5 10MegΩ, R2 5 10MegΩ, RB 5 4.7MegΩ, RL 5 12KΩ, IC 5 0.3375 mA

IC 5 0.0003375 A, or IC 5 337.5 µA

Q1 5 2N3904, BT 5 9 volts

G ~ –145 as measured with 20 mV peak to peak sinewave at Vin with a 50Ω source resistance generator.

G 5 –144 as calculated. That is, G 5 –(0.0003375A/0.026 v) 12KΩ ~ –144.

NOTE: 12KΩ || 10MegΩ = 12KΩ within about 0.12%.

See below for the input and output signals.

FIGURE 8-12 Amplifier 20 mV peak to peak signal input on top trace. Bottom trace shows a 2.7-volt peak to peak (inverted) output signal with 10 percent second harmonic distortion; and notice the rounder positive cycle and narrower negative cycle.

We can mitigate the distortion problem by using a series resistor, R_ser, as shown in Figure 8-13.

FIGURE 8-13 Lower distortion for the same output swing.

Typically, if we want lower distortion from an amplifier, we have to give up some of the voltage gain.

For example, if the gain of the amplifier is about 145, we can have lower distor- tion for the same output voltage if we set the series base resistor at about 10 3 R_in base. Again, if we use the previous example with:

R1 5 10MegΩ, R2 5 10MegΩ, RB 5 4.7MegΩ, RL 5 12KΩ, IC 5 0.3375 mA or

IC 5 337.5 μA

Q1 5 2N3904, BT 5 9 volts

IB is then calculated as about (9 v – 0.7 v) 4.7MegΩ 5 1.766 μA. We can now find β 5 IC/IB 5 (337.5 μA/1.766 μA):

β 5 191

This leads to R_in base 5 β(0.026 v/IC) or R_in base 5 191(77Ω).

R_in base 5 14.7KΩ. If we make R_ser about 10x of R_in base, then R_ser ~ 150KΩ. The resulting distortion can now be compared with R_ser 5 0 Ω and R_ser 5 150KΩ for gain and distortion at Vout for the same output voltage. See Table 8-1 that shows the effect on distortion with different values for input series resistor, R_ser. There is a trade-off in that you can have lower distortion but at the expense of lower voltage gain.

TABLE 8-1  Distortion Measurements for the Same Amplitude Output at Vout

R_serVout pk to pkVin pk to pkGainVout Distortion
~ 0 Ω5.18 volts p-p38.9 mV p-p–133~ 20%
150KΩ5.18 volts p-p429 mV p-p–12~ 2%

NOTE that with the gain reduced by about tenfold, the distortion is reduced by ~10 as well. Also it is preferred that RB >> R_ser for reducing distortion.

See Figures 8-14 and 8-15, where the top trace waveforms are the input signals, and the bottom traces show the output signals. Note the phase inversions between output and input signals.

FIGURE 8-14 Output waveform on the bottom trace with R_ser = ~ 0 Ω, and note the compression on the positive cycle and narrowing on the negative cycle that denotes 20 percent harmonic distortion.

Figure 8-15 shows when a “linearizing” series 150K resistor is added.

FIGURE 8-15 Output waveform on bottom trace with R_ser = 150KΩ, which shows almost no compression or narrowing distortion compared to Figure 8-14. Harmonic distortion was measured at approximately 2 percent.

Output Swing Determined by IC and RL || R2

To maximize output voltage swing, we should bias the collector current, IC, such that the DC collector voltage is in a range of 40 to 60 percent of the supply voltage. For example, if the supply voltage such as BT2 in Figure 8-13 is 10 volts (e.g., 8 AA rechargeable 1.25-volt batteries in series), then a good starting point will be at the 50 percent of 10 volts or 5 volts DC at the collector of Q1. RB would be selected for this. The reason we want the DC collector voltage close to 50 percent of the supply voltage is to provide maximum output AC voltage swing. However, depending on the appli- cation, sometimes we just need a few volts of AC swing, peak to peak.

The maximum peak voltage swing without distortion is calculated as:

Vout max peak to peak 5 2 3 IC 3 RL||R5, where VC ≥ 50% of power supply volt- age

For example, suppose you have BT2 5 6 volts and RL 5 3KΩ, and you bias RB such that:

IC 5 1 mA DC

This means that VC 5 6 volts – IC(RL) 5 6 volts – 1 mA (3KΩ) 5 6 volts – 3 volts

or VC 5 3 volts.

VC 5 3 volts, which is 50 percent of BT2 5 6 volts. This is a good starting point. If R5 is >> RL, such as RL 5 1MegΩ, then the output AC voltage swing will be close to 6 volts peak to peak.

This is because with R5 >> RL, RL || R5 ~ RL:

Vout max peak to peak 5 2 3 IC 3 RL||R5 ~ 2 3 IC 3 RL 5 2 3 1 mA 3 3KΩ 5

2 3 3 volts

Vout max peak to peak ~ 6 volts peak to peak

However, if R5 5 RL, then RL || R5 5 0.5 RL and the voltage swing will be reduced by 50 percent. For example, when RL 5 3KΩ and R5 5 3KΩ, the output voltage swing is reduced by 50 percent.

Vout max peak to peak 5 2 3 IC 3 RL||R5 5 2 3 IC 3 RL||RL 5

2 3 IC 3 0.5 RL 5 IC 3 RL 5 Vout max peak to peak

Vout max peak to peak 5 1 mA 3 3KΩ 5 3 volts peak to peak

If the 3 volts peak to peak Vout max is sufficient, then you do not need to change the circuit. However, if R5 is lower in value such as 2KΩ or 1KΩ and you still want 3 volts peak to peak output, then you have to make RL a smaller resistance such as lowering it to 2KΩ or 1KΩ while reselecting RB to have VC 5 3 volts DC such that the DC collector current, IC is increased.

So far, the one-transistor amplifier circuits presented have many variables in terms of gain, input resistance, and DC operating points such as collector current IC and collector voltage VC due to the variation in β, the current gain. If you need to build many of these amplifiers and make them repeatable, you can buy the transis- tors in tape form as shown in Figure 8-16. They should be reasonably matched in terms of turn-on voltage and β, but check them with a DVM for confirmation.

FIGURE 8-16  Transistors on tape are reasonably matched for b and VBE turn-on voltage.

Troubleshooting the One-Transistor Amplifier

The one-transistor amplifier may not work for the following reasons:

•     If the transistor is put in backwards with the collector and emitter swapped, then the collector current will be very low because β  1 or 2. This will result in very low gain since the collector current will be too small. So, expect VC to be almost the same voltage as the power supply, which says that the voltage across RL is close to zero volts.

•     If the collector and emitter terminals are reversed, and the power supply voltage is > 6 volts DC, there’s a good chance that the transistor will act like a Zener diode, and the voltage at the collector will be in the range of 5 or 7 volts DC, but there will be very little signal output at Vout.

•     If you replace the transistor, but find that the output signal is low, check the DC collector voltage, VC. If the collector’s voltage VC is close to 0 volts (e.g.,

< 0.5 volt), the transistor is in saturation because there is too much base current. Reselect the base driving resistor, RB, to have VC at about half the supply voltage, or select a transistor with lower current gain, β. You may also have to confirm the proper resistance values for RL and RB. If RL is too high in resistance and or if RB is too low in resistance, it can cause the transistor to be in saturation.

•     If you think you connected everything correctly but you find strange DC readings and output signals, make sure the transistor is the correct polarity (e.g., NPN or PNP). Sometimes it’s easy to inadvertently put in the wrong polarity transistor and/or put it in with swapped leads.

•     Check power supply voltage, and always add a bypass capacitor such as 0.1 μf to 1 μf across the power supply voltage source close to the transistor amplifier. Also note the supply voltage and choose the bypass capacitor to have at least twice the voltage rating. For example, if the supply voltage is 12 volts, choose a 25-volt to 100-volt capacitor.

Using Negative Feedback to Build “Mass Production” Amplifiers

If we look at the one-transistor amplifier schematic in Figure 8-13 with its output waveform in Figure 8-14, we will notice that the AC signal going into the base pro- duces a signal at the collector that is in negative phase or opposite phase of the input. This makes sense because the DC collector voltage, VC BT2 – IC (RL). Since

IC is proportional to base current, IB, and IB is some function of the base voltage, then an increase in base current causes an increase in collector current. But an increase in collector current, IC, causes a decrease in VC due to VC 5 BT2 – IC (RL). For example, if BT2 5 10 volts, RL 5 1KΩ, then VC 5 5 volts – IC 1KΩ. If we start out with IC 5 1 mA, then VC 5 5 volts – 1 mA 1KΩ 5 5 volts – 1 volt.

VC 5 4 volts

Suppose we increase the collector current such that IC  3 mA, then: VC 5 5 volts – 3 mA 1KΩ 5 5 volts – 3 volts 5 2 volts

VC 5 2 volts

As we can see, an increase in collector current results in a decrease in collector voltage. Since an increase in base current causes an increase in base voltage, then it follows that an increase in base voltage causes an increase in collector current, which causes a decrease in collector voltage. Thus, we have a “negative” phase relationship between the base and collector terminals. If we have this relationship, we can apply a self-biasing resistor between the base and collector as shown in Figure 8-17. If this resistor is chosen properly, we can set a DC collector voltage and collector current that is not as sensitive to β variations. As long as β >> 1 such as β ≥ 20, the DC col- lector current that is set by this circuit will not vary much whether the transistor has a β of 50 or a β of 500. Also, as long as the AC signal’s gain is kept to ≤ 10, the gain will be set by two resistors and it again will be insensitive to β variations.

FIGURE 8-17 A simple self-biasing one-transistor amplifier via R_BC for “insensitivity” to b.

DC Analysis of Self-Biasing Amplifier

With a negative feedback resistor R_BC that is connected from the collector to base, a DC voltage is established at VC. For the DC analysis, see Figure 8-18.

Typically, the base-emitter voltage at VB is ~ 0.7 volt. If Q1’s β is greater than 50, then there’s a good chance that Q1’s base current, IB, is very small such that there is a very small voltage across R_BC, V_R_BC 5 IB 3 R_BC. (Refer to Figure 8-18.)

FIGURE 8-18 DC analysis circuit for the self-biasing one-transistor amplifier via removing the DC blocking capacitors.

NOTE that VC = V_R_BC + VB or VC = (IB x R_BC) + VB.

If this voltage across resistor R_BC, V_R_BC << 0.7 volt, then VC ~ VB. A com- monly used transistor, such as2N4124 or equivalent, generally has β ≥ 50 at currents from 100 μA to about 20 mA.

For power supply voltages that are >> 0.7 volts (e.g., ≥ 5 volts) even with base current IB flowing through R_BC in the order of V_R_BC 5 0.7 volts such that VC 5 1.4 volts versus VC 5 0.7 volts, the collector current will vary < 25 percent for BT2 ≥ 5 volts.

For example, with BT2 5 5 volts, IC ~ (BT2 – VC)/RL.

With VC 5 0.7 volt, IC ~ (5 volts – 0.7 volt)/RL or IC0.7v ~ 4.3 volts/RL. With VC 5 1.4 volts, IC ~ (5 volts – 1.4 volts)/RL or IC1.4v ~ 3.6 volts/RL.

We can take the ratio of the two collector currents for VC 5 0.7 volt and 1.4 volts

to determine the change in collector current.

(IC0.7v /IC1.4v) 5 [(4.3 volts/RL)/(3.6 volts/RL)], the RL’s cancel out, leaving (IC0.7v /IC1.4v) 5 (4.3 volts/ 3.6 volts) 5 1.1944, which is a 19.44 percent change in collector current.

Let’s take a look at some typical component values. Q1 5 2N4124 or 2N3904, BT2 5 5 volts, RL 5 3300Ω, R_BC 5 100KΩ, with β 5 170, the measured VC 5 1.3 volts. The collector current with β 5 170 is:

IC ~ (5 volts – 1.3 volts)/3300Ω 5 1.12 mA 5 IC

If we replace Q1 with a 2N5089 where β 5 461, VC 5 0.9 volt, and IC 5 (5 volts – 0.9 volts)/3300Ω 5 1.24 mA for β 5 461, which is only about a 10.8 per- cent change with 1.12 mA for β 5 171. So even when β increased from 170 to 461, a 2.77-fold increase, the collector current, IC, only increased by about 10.8 percent.

Having VC biased at a DC voltage of about 1 volt will limit the voltage swing from about 0 volt to about 2 volts or 2 volts peak to peak before clipping of the output waveform occurs.

To maximize output voltage swing, it is generally better to have VC set to about one half the supply voltage. For example, if BT2 5 5 volts, and VC is set to 2.5 volts DC, then the output swing can be from about 0 volts to nearly 5 volts, or close to 5 volts peak to peak. By adding a resistor (R_BE) across the base-emitter terminals, we can raise VC’s DC voltage. See Figure 8-19.

FIGURE 8-19  Adding an extra resistor R_BE to form a voltage divider with R_BC.

By adding R_BE, we can raise the DC collector voltage. See Figure 8-20 for the DC analysis schematic.

FIGURE 8-20  DC analysis circuit for Figure 8-19.

Since VB ~ 0.7 volt, we can work backwards to find VC since the voltage across R_BE is ~ 0.7 volt. By using a voltage divider formula and neglecting the base cur- rent from Q1, we have:

0.7 volt 5 VB 5 VC [R_BE/(R_BE 1 R_BC)]

0.7  volt/[R_BE/(R_BE 1 R_BC) ] 5 VC VC 5 0.7 volt [(R_BE 1 R_BC)/R_BE]

In essence the DC voltage at the collector, VC, is a scaled “up” voltage of VBE. For example, if we want VC 1.4 volts, then R_BC 5 R_BE.

VC 5 0.7 volt [(R_BE 1 R_BE)/R_BE ] 5 0.7 volt [2R_BE/R_BE ] 5 0.7 volt 3 2

VC 5 1.4 volts

A general formula where it is easier to just find the ratio of R_BC to R_BE based on having a specified VC is shown here:

R_BC/R_BE 5 (VC/0.7 v) – 1

For example, if BT2 5 5 volts, RL 5 3300Ω, and we want VC 5 2.5 volts, then: R_BC/R_BE 5 (2.5/0.7 v) – 1 5 3.57 – 1 5 2.57

R_BC 5 2.57 (R_BE)

We can make R_BE 39KΩ so that R_BC 5 2.57 (39KΩ)

R_BC 5 100KΩ

In general VC will be a bit higher than calculated using a general-purpose tran- sistor such as with a 2N4124 transistor. With a higher β transistor such as a 2N5089, the calculated VC voltage will be closer in practice.

To preserve good voltage swing, R_BC >> RL. For example, if RL 5 1KΩ, R_BC

≥ 10KΩ. However, we will see that the simple 1 transistor amplifier has a “drawback” in terms of input resistance being too low. But not to worry—we use the lower input resistance to our advantage by adding a series input resistor to set the gain of the amplifier (see R1″ in Figure 8-25 if you are curious)..

AC Analysis of a Self-Biased Amplifier

Adding the feedback resistor R_BC causes a lower total input resistance as “seen” by Vin (see Figure 8-21).

FIGURE 8-21  AC analysis circuit for the one-transistor feedback amplifier.

With negative feedback via the collector to base resistor, R_BC, this resistor and with the amplifier’s gain, G, will result in a low-value resistor, Ri, referenced to ground, where Ri is in parallel with R1 and R_BE.

Ri ~ [R_BC/(1 – G)] || R_in base, where G 5 Vout/Vin, and since this is an inverting amplifier, G will be a negative number.

G ~ –(IC/0.026 v) 3 (RL || R2 || R_BC)

The gain, G, is calculated to be the same as the simple transistor amplifier shown in Figure 8-10 when R_BC >> RL || R2. For a good enough approximation where R_BC ≥ 10 3 (RL || R2), use:

G ~ –(IC/0.026 v) 3 (RL || R2)

For example, if the supply voltage BT2 5 12 volts, Q1 5 2N5089 where β 5 400, RL 5 4700Ω, R_BC 5 100KΩ, R_BE 5 20KΩ and VB 5 0.7 volt,

then VC 5 0.7 v [(R_BC/R_BE) 1 1].

VC 5 0.7 volt [(100K/20K) 1 1] 5 0.7 volt [6]

VC 5 4.2 volts

IC ~ (BT2 – VC)/RL 5 (12 volts – 4.2 volts)/4700Ω 5 7.8 volts/4700Ω

IC ~ 1.66 mA 5 0.00166 A

Let R2 5 1MegΩ so that RL || R2 5 4700Ω || 1MegΩ ~ 4700Ω 5 RL || R2.

G ~ –(IC/0.026 v) 3 (RL || R2)

G ~ –(0.00166 A/0.026 v) 3 (4700Ω) G ~ –300

R_in base 5 β(0.026 v/IC) 5 400(0.026 v/0.00166 A)

R_in base 5 6.26KΩ 5 internal resistance across the base and emitter of Q1 Ri ~ [R_BC/(1 – G)] || R_in base

Ri ~ [100KΩ/(1 – –300)] || 6.26KΩ 5 [100KΩ/(1 1 300)] || 6.26KΩ Ri ~ 332Ω || 6.26KΩ

Since 332Ω << 6.26K, 332Ω || 6.26KΩ ~ 332Ω, which leads to an approximation


Ri ~ 332Ω

R_in base is the resistance into the base-emitter junction of the transistor that does not include any of the other external resistors such as R_BC, R_BE, and R1.

The resistance [R_BC/(1 – G)] is due entirely to Vin’s signal current flowing through R_BC, which results in a lower equivalent resistor, R_BC/(1 – G) referenced to ground. So, the question arises as to how we get such a lower equivalent resistor that is related to the amplifier’s gain, G? See Figure 8-22.

FIGURE 8-22  Input and output waveforms showing increased AC voltage across R_BC.

The amplifier produces an amplified (e.g., a larger amplitude) signal at the out- put, Vout, which is in opposite phase of the smaller amplitude input signal, Vin. This opposite phase signal at Vout then pulls extra current through the resistor R_BC. By pulling extra current, it makes it appear that Vin is driving an equivalently lower- value resistor that is referenced to ground like R1.

An example of this lowered resistance effect is if we imagine that the gain, G 5

–1 and R_BC 5 1KΩ, then what happens when Vin 5 11 volt? The output voltage is then – 1volt due to G 5 –1. However, the current flowing through the 1KΩ has a potential difference of 11 v – –1v or 2 volts. This means the resistor current is now 2 v/1KΩ 5 2 mA. A 11-volt input across an equivalent resistor referenced to ground that will drain 2 mA would result in a 500Ω resistor since 1 v/500Ω 5 2 mA.

In general, the voltage across resistor R_BC is V_R_BC, which has Vin on one side and Vout on the other side. Thus, V_R_BC 5 Vin – Vout, but Vout 5 –G Vin, so:

V_R_BC 5 Vin – –GVin 5 Vin 1 G Vin V_R_BC 5 Vin (1 1 G)

The current through R_BC is then:


and by substituting Vin (1 1 G) for V_R_BC:

I_R_BC 5 Vin (1 1 G)/R_BC

We want to now model an equivalent resistor referenced to ground (as shown in Figure 8-23) that will drain the same amount of current as I_R_BC, which is I_R_BC 5 I_equiv 5 Vin(1 1 G)/R_BC.

FIGURE 8-23 Equivalent resistor referenced to ground that drains the same current as I_R_BC.

Let’s see if the current, I_equiv, is the same as I_R_BC.

R_BC/(1 1 G)               R_BC 

I_equiv 5    Vin     Vin (1 1 G) 5 Vin (1 1 G)/R_BC 5 I_R_BC

I_equiv 5 I_R_BC

This confirms that the equivalent resistor referenced to ground is R_BC/(1 1 G). This makes sense because it would take a smaller-value resistor referenced to ground with just Vin applied to it to drain the same amount of (higher) current that

would have to be a resistor divided in value by (1 1 G). See Figure 8-24.

Note if the amplifier has no gain where G 5 0 that causes Vout 5 0 volt, then essentially R_BC is shorted to ground on the collector of Q1, and we do indeed have R_BC as the equivalent resistor reference to ground.

R_BC/(1 1 G)

with G 5 0

R_BC/(1 1 0) 5 R_BC

FIGURE 8-24  AC model where Ri = [R_BC/(1 – G)] || [R_in base].

The output resistance, Ro, is actually dependent on the input signal’s series source resistance. If Vin is driven by a pure voltage source or very low output imped- ance amplifier, Ro 5 RL||R2||R_BC, which is similar to the simple amplifier in Figures 8-3 and 8-7. Normally, R_BC >> RL||R2, so Ro ~ RL||R2.

Because R_BC lowers the overall input resistance seen by Vin to almost “short” circuit compared to the other resistors, R1, R_BE, and R_in base, generally, we do not drive the input directly with a signal source with low source resistance.

However, we can improve upon the amplifier in Figure 8-19 by adding a series resistor that will lower the gain, Vout/Vin, but then improves on raising the input resistance, lowering the output resistance, and also lowering the distortion for the same amplitude output. See Figure 8-25.

FIGURE 8-25 A modified one-transistor feedback amplifier with a series input resistor, R1″.


We can define the gain G’ 5 Vout 5 –(IC/0.026 v) 3 (RL || R_BC) that is relative to the signal voltage at the base of Q1, VB and Vout, and not taking R2 into account

~ – 

for now.

And then we can define the actual gain

Vout Vin

R_BC R1″

where –

R_BC R1″

≤ 5% of G’.

Output Resistance Ro’

The reader can skip over this section pertaining to the output resistance, Ro’, if desired. These calculations are long. The gist is having R_BC as a feedback resistor results in a lower output resistance than RL||R2.

The output resistance, Ro’, not including R2 is approximately:

Ro’ 5 {[(R_BC 1 R_in base || R_BE || R1″)/(R_in base || R_BE || R1″)] 3 (0.026 v/ IC)} || RL

Here are a couple of calculations with their measured results.

With BT2 5 5 volts, RL 5 2700Ω, R_BC 5 100KΩ, R_BE 5 40.2KΩ, and

R1″ 5 20KΩ, the calculated gain (Vout/Vin) 5 –(R_BC/R1″) 5 –100K/20K. Vout/Vin 5 –5.0

Measured gain is –4.5, within 10 percent of the expected gain of –5.

Now let’s calculate the output resistance, with R2 removed from the circuit in Figure 8-25.

The calculated VC ~ VBE (1 1 R_BC/R_BE) where VBE ~ 0.7 v:

VC ~ 0.7 v(1 1 100K/40K) 5 0.7 volt (3.5)

VC ~ 2.45 volts. The measured VC was 2.3 volts.

With BT2 5 5 volts and VC 5 2.45 volts with RL 5 2.7KΩ, the calculated IC ~ (BT2 – VC)/RL since R_BC 5 100KΩ >> RL 5 2.7KΩ:

IC ~ (5 v – 2.45 v)/2.7KΩ IC ~ 0.907 mA

The current gain β ~ 450 for the 2N5089.

R_in base 5 β (0.026 v/0.000907A) 5 450(286Ω)

R_in base 5 12.9KΩ

With R_BC 5 100KΩ, R_in base 5 12.9KΩ, R_BE 5 40.2KΩ, R1″ 5 20KΩ, and RL 5 2700Ω

Ro’ ~ {[(R_BC + R_in base || R_BE || R1″) /(R_in base || R_BE || R1″)] 3 (0.026 v/ IC)} || RL

where R_BC in parallel with RL is neglected since R_BC 5 100KΩ >> RL 5 2.7KΩ. Therefore:Ro’~{[(100KΩ1(12.9KΩ||40.2KΩ||20KΩ)/(12.9KΩ||40.2KΩ||20KΩ)]

3 (0.026 v/0.000907 A)}|| 2.7KΩ

NOTE: (12.9KΩ||40.2KΩ||20KΩ) = 6.56KΩ.

Ro’ ~ {[(100KΩ 1 6.56KΩ)/(6.56KΩ)] 3 (0.026 v/0.000907 A)}|| 2.7KΩ Ro’ ~ {16.24 3 (0.026 v/0.000907 A)}|| 2.7KΩ

Ro’ ~ {16.24 3 (28.66Ω)}|| 2.7KΩ Ro’ ~ 465Ω|| 2.7KΩ

Ro’ ~ 465Ω|| 2.7KΩ

Ro’ ~ 397Ω calculated

Ro’ ~ 390Ω measured via setting R2 5 390Ω and noticing that the signal dropped by 50 percent.

The output resistance, Ro’, calculation is rather long, and sometimes we do not have time to do this. Here are some rules of thumb concerning the transistor amplifier.

1.            Use low-value gains that are typically less than 5 or 10, such that R_BC/R1″ ≤ 10.

2.            Make sure that R_BC >> RL by at least tenfold. Thus, R_BC does not lower the gain G by much since as “seen” by the collector of Q1, RL and R_BC are essentially in parallel. The reason is that the resistors R_BE and R_in base that are coupled to ground and connected to R_BC at the base are << R_BC, that is (R_BE || R_in base) << R_BC.

3.            Bias the DC voltage at the collector, VC, to be at one-half the power supply voltage.

4.            Use as high β transistor as you can, such as a 2N5089. If you use a lower β transistor such as a general-purpose type, 2N3904, then you may have to set for lower gains such as R_BC/R1″ ≤ 5.

5.            Generally, bias the collector current, IC, in the range of 0.2 mA to 5 mA with the appropriate value RL resistor that is scaled from the example at ~ 1 mA where RL 5 2700Ω. For instance, if you are running IC at 200 μA, then RL ~ 5 3 2700Ω or about 12KΩ, and R_BC ~ 5 3 100KΩ 5 510KΩ, R_BE ~ 5 3 40.2KΩ 5 200KΩ, and R1″ 5 100KΩ. Going the other direction with IC 5 5 mA, all the resistors that were used for IC ~ 1 mA are

scaled by (1/5) or 20 percent or 0.2. Thus, RL ~ 0.2 3 2700Ω ~ 560Ω, R_BC ~ 0.2 3 100KΩ ~ 20KΩ, R_BE ~ 0.2 3 40.2KΩ 5 8.2KΩ, R1″ ~ 0.2 3 20KΩ ~ 3.9KΩ.

Note the input capacitor, C1, can be scaled according to the collector cur- rents. For example, if a 1 μf is used for 1 mA collector, then 4.7 μf is used for 5 mA IC, and 0.22 μf is used for 0.2 mA collector current. However for C1, you can always just use the highest value capacitor for all cases such as 4.7 μf or greater capacitance. The output capacitor C2 should be greater than or equal to 4.7 μf to ensure good low frequency response.

6.            One feature of this amplifier is that the performance increases as you set the gain lower via lower ratios of R_BC_ to R1″. This is normally done by increasing the value of R1″. The increase in performance parameters result in lower output resistance, Ro’, and lower harmonic distortion at Vout for the same output amplitude.

7.            Another feature of this amplifier is that you can set the inverting gain to attenuate the signal. That is, 0 < |R_BC/R1″| < 1. For example, if R_BC 5 100KΩ, we can set R1″ 5 300KΩ for a gain of – (1/3) or – 0.33. This can be useful if the amplifier is interfacing with a signal source whose amplitude exceeds the amplifier’s power supply voltage, BT2. For example if your generator or signal source provides 10 volts peak to peak, and your amplifier can only deliver about 4 volts peak to peak, having the gain set to

–0.33 will keep the output from clipping since the output voltage will be 10 volts 3 (–0.33) peak to peak or 3.3 volts peak to peak.

Referring to #6, see Figure 8-26, which shows the output waveform for the cir- cuit in Figure 8-25, where the gain is ~ –94 via input series resistor R1″ 5 25Ω. In comparison see Figure 8-27 where the distortion is lower for the same output voltage because R1” is increased to 20KΩ for a (lower) gain of – 4.2.

To lower output signal distortion R1″ is increased to 20kΩ. See Figure 8-27 bot- tom trace.

FIGURE 8-26 R1″ = 25Ω, Vin = 32 mV peak to peak (top waveform), and bottom waveform Vout = 3 v peak to peak with 12 percent harmonic distortion. Gain ~ –94 = Vout/Vin. Also note the phase inversion between the output signal Vout and the input signal Vin.

FIGURE 8-27 R1″ = 20KΩ, Vin = 720 mV peak to peak (top trace), and Vout = 3 v peak to peak for a gain of – 4.2 (bottom trace). Harmonic distortion measured at 1 percent. Again, note the phase inversion of Vout with respect to Vin.

Another Common Emitter Amplifier

We will now look at another type of common emitter amplifier with a series emitter resistor derived from an LED drive circuit. See Figure 8-28, which shows a constant current source LED drive circuit, and then Figure 8-29, which converts the LED drive circuit into an amplifier.

FIGURE 8-28  A constant current source amplifier where the LED drive current = VE/RE”.

We can reconfigure the constant current source circuit to an amplifier by replac- ing the LED with a load resistor, RL, and coupling an input signal voltage to the base with a capacitor. See Figure 8-29.

FIGURE 8-29 A simple common emitter amplifier with a series emitter resistor RE”.

To bias this correctly, generally “Vref” 5 BT2[R4/(R3 1 R4)] is less than half the supply but ≥ 0.5 volts. Typical collector currents can range from 50 μA to about 20 mA for a small signal transistor such as the 2N3904. Generally, the voltage divider resistors ≥ 470Ω, and the emitter resistor RE” that sets the DC collector current can be anywhere from about 100Ω to 10KΩ. Collector current IC 5 VE/RE”, where β >> 1, which is usually the case. In terms of gain, a first approximation has Vout/Vin ~ – RL/(re 1 RE”), where re 5 (0.026 v)/IC. Again, this is an inverting amplifier. Generally, it’s harder to provide very high gains from this configuration unless very high supply voltages are used. For example, if we want a gain of –100, and IC 5 1 mA, with VE 5 1 volt and RE” 5 1000Ω, then RL has to be ~100KΩ. This would require BT2 >100 volts. For example, BT2 5 200 volts so that VC is operating at 100 volts DC. Also, a general-purpose transistor will have insufficient breakdown voltage and a high-voltage transistor such as a 2N3439 or MPSA42 would be used instead. If we want instead a gain ~ –10, then we can have RL 5 10KΩ, BT2 5 20 volts, Q1 5 2N3904, with VC ~10 volts when IC 5 1 mA.

An emitter bypass capacitor CE2 provides higher gain (Vout/Vin) at the expense of higher distortion. But this circuit allows for lower supply voltages. See Figure 8-30.

FIGURE 8-30  Capacitor CE2 effectively AC grounds Q1’s emitter.

By adding a resistor (RE2) in series with capacitor CE2, the AC gain can be set without upsetting Q1’s DC bias points (e.g., VB and IC). See Figure 8-31, which includes RE2.

FIGURE 8-31 A second series emitter resistor, RE2, allows for increasing the gain without changing the DC bias conditions. Capacitor CE2 has a large capacitance such that it is an AC short circuit.

In Figure 8-31, with RE2 in series with CE2, the gain G 5 –[RL/(re 1 RE” || RE2)], where re 5 (0.026 v/IC). To find the DC operating points, VC, IC, and VE, we remove the capacitors as shown in Figure 8-32.

FIGURE 8-32  A DC analysis circuit of Figure 8-31 where capacitors are removed.

With the capacitors removed and neglecting DC base currents due to IB, VB ~ BT2 [R4/(R3 1 R4)] and with Q1’s turn-on voltage, VBE ~ 0.7 volt DC, VE ~ VB – 0.7 volt. The collector current IC ~ IE, the emitter current when β >>1, which is usually the case. We now have:


We now have to also find AC resistance, R_in base, which can be found via IC, β, re, RE”, and RE2.

R_in base 5 β re 1 (β 1 1) (RE” || RE2), where re 5 (0.26 v/IC).

For the AC analysis model, see Figure 8-33.

FIGURE 8-33 An AC signal analysis circuit.

For example, if BT2 5 6 volts, R3 5 40.2KΩ, R4 5 20KΩ, RE” 5 3900Ω, RL 5 10KΩ, RE2 5 1KΩ, β 5 150, then VB ~ 6 v (20K/(40.2K + 20K) ~ 6 v (20/60.2K)

VB ~ 2 volts.

VE ~ VB – 0.7 v 5 (2 – 0.7) v or VB 5 1.3 volts. IC ~ VE/RE” 5 1.3 v/3900Ω or IC ~ 0.00033A.

To find the gain, G, we calculate for:

re 5 (0.026 v/IC) 5 (0.026 v)/(0.00033A) or re 5 78.8Ω G 5 –[RL/(re 1 RE” || RE2 )

RE” || RE2 5 3900Ω || 1KΩ 5 796Ω, thus

G 5 –[10K/(78.8 1 796)] or G ~ –11.4 , and with β 5 150 we have:

R_in base 5 β re + (β 1 1) (RE” || RE2) 5 150 (78.8Ω) 1 (150 1 1) (796Ω)

R_in base ~ (11.82KΩ 1 120.2KΩ) R_in base 5 132.02KΩ

When RE” || RE2 >> re, the emitter resistors RE” and RE2 are the main contribu- tors for R_in base that leads to: R_in base ~ (β 1 1) (RE” || RE2).

From Figure 8-34, to find the AC input resistance to the amplifier, it is (R_par || R_in base’), where R_par 5 R3 || R4. With the example where R3 5 40.2KΩ, R4 5 20KΩ, and the calculated R_in base’ 5 132.02KΩ, the input resistance is:

R3 || R4 || R_in base’ 5 (40.2KΩ || 20KΩ) || 132.02KΩ 5 (13.33KΩ) || 132.02KΩ

R3 || R4 || R_in base’ 5 12.13KΩ

FIGURE 8-34  AC signal analysis block diagram with output resistance Ro = RL.

The input resistance 5 12.13KΩ.

We can reiterate the gain calculation for G from knowing the collector current and the values of the collector resistor RL and emitter resistors, RE” and RE2 along with re.

G 5 –[RL/(re 1 RE” || RE2 )

re 5 78.8Ω 5 (0.026 v/IC), where IC 5 0.00033 A as calculated previously.

RE” || RE2 5 3900Ω || 1KΩ 5 796Ω, RL 5 10KΩ and thus G 5 –[10K/(78.8 1 796)]

G ~ –11.4

The output resistance, Ro 5 RL, and with RL 5 10KΩ in this example, Ro 5 10KΩ.

NOTE: An amplifier having a high output resistance such as 10KΩ results in a gain reduction of G when driving a subsequent device (e.g., input stage of another amplifier). If Vout is loading into another device that has an effective 10KΩ to ground will cause the gain, G, drop from –11.4 to half or G  –5.7.

Typical input resistances of audio power amplifiers are between 1KΩ to 100KΩ. For example, an amplified computer stereo loudspeaker with a 3.5 mm connec-

tor has a typical input resistance of 10KΩ.

Troubleshooting the Amplifier in Figure 8-31

1.            For measuring the DC conditions, turn off the AC signal source and measure the collector, base, and emitter DC voltages, VC, VB, and VE. Make sure that the transistor is in the amplifying region (forward active region) and not in the saturation region. This can be confirmed by measuring with a DVM that

the voltage across the collector and emitter (VCE) > 11 volt DC, or that the collector to base (VCB) voltage > 0.7 volt. Generally, we should expect that VCE or VCB is in the range of at least a quarter of power supply voltage. For example, if BT2 5 12 volts, then VCE ≥ 3 volts. In another example, if BT2 5 5 volts, then the DC voltages across the collector to base (VCB) and across the collector to emitter (VCE) should be ≥ 25 percent (5 volts) or ≥ 1.25 volts.

2.            Confirm that R3, R4, RE”, and RL are correct. Measure the resistors with the power turned off and with one lead of each resistor disconnected from the circuit. In a special case, you can measure RE2 in circuit because it has a series DC blocking capacitor CE2, but you may have to keep the ohm-meter’s probes across RE2 for about 10 seconds to let the resistance measurement settle.

3.            With power turned back on, confirm that the voltages comply with the expected voltage from the voltage divider circuit with R3 and R4. That is, VB 5 BT2[R4/(R3 1 R4)], and VE 5 VB – 0.7 volt.

4.            Confirm that the emitter current is approximately equal to the collector current by measuring the voltage across RL 5 VRL and voltage across RE” 5 VRE. Then confirm via calculation from the resistance values of RL and RE” and the measured voltages, VRL and VRE that VRL/RL ~ VE/RE” is within 15 percent if 5 percent resistors are used.

5.            If the DC voltages seem unstable, the amplifier may be oscillating at >100MHz. Try inserting a 100Ω series base resistor close to Q1 and see if the DC voltages stabilize and measure to their expected values.

6.            With the signal generator turned on having a small amplitude such as 100 mV peak to peak at Vin, measure the AC signal at VE with an oscilloscope. The AC signal at VE should be approximately 100 mV peak to peak within about 25 percent on the low side (e.g., VE’s output AC voltage is in the range of 75 mV to 100 mV peak to peak). Note that the AC voltage at the emitter is the same phase as the base since the emitter terminal acts like having a 0.7 volt DC source voltage between the base and the emitter. A DC voltage source such as the VBE turn-on voltage in series with an AC signal source at the base cannot change the phase of the AC signal at the emitter. The output signal at the collector, VC, should be G 3 100 mV peak to peak, or in this example, where G 5 –11.4, Vout AC should be about 1.14 volts peak to peak that is out of phase with the input signal at the base.

7.            In Figures 8-30 and 8-31 emitter capacitors CE and CE2 are chosen to determine the low-frequency response. The 10 μf input and output capacitors, C1 and C3, are designed with sufficient capacitance for most applications with a 20 Hz or lower cut-off frequency. For example, for audio applications, where a low-frequency response of ≤ 20Hz is required, C1 and C3 generally will work because input and output resistances are generally greater than 1KΩ.

For example, the cut-off frequency with a 10 μf capacitor and 1KΩ resis- tor is:

1/[2π 10 uf (1KΩ)] ~ 16 Hz

To determine the worst-case capacitance values for CE and CE2, choose the low-end frequency you want such as 20Hz. Then use the following for- mula:

CE or CE2 5 1/[2π fc re]

Let fc 5 20 Hz 5 low-frequency cut-off frequency for an audio applica- tion that responds to bass note frequencies.

For example, if IC 5 1 mA, then re 5 (0.026 v/0.001A) or re 5 26Ω.

CE or CE2 5 1/[2π 20Hz 26Ω ] 5 3.06 3 10–4 Farad or 306 μf

The closest commercially available value is 330 μf, but to ensure a safety margin just in case, use a 470 μf unit if possible. Since the emitter voltage is generally less than 25 volts, you can use at least a 16-volt electrolytic capacitor. Make sure the (–) terminal of CE or CE2 is grounded.

8.            Confirm that any electrolytic capacitors in the circuit are correctly connected (polarity-wise) to avoid reverse biasing. Reverse biasing electrolytic capacitors usually cause the DC bias points to be wrong, due to current leakage through the electrolytic capacitors.

9.            The common emitter amplifier in Figure 8-30 where capacitor CE is an AC short circuit from emitter to ground has the same input amplitude limitation as the simple bias circuit in Figure 8-11 for tolerable distortion at the output, Vout. By tolerable distortion, an example would be for intelligible voice signals, < 10 percent harmonic distortion is workable. That is, generally the input signal, Vin, is limited to less than 10 mV or 20 mV peak to peak when driven by a low impedance source signal such as a 50Ω generator. If Vin is driven with a larger value source resistance such as a ≥ 10KΩ series resistor, the distortion will be lower but the voltage gain will be lower as well due to the input resistance forming a voltage divider circuit.

For higher-fidelity applications that require lower distortion at the output less than 1 percent, use the circuit in Figure 8-31. See #10 below concerning distortion calculations.

10.            In Figure 8-31 where CE2 is an AC short circuit (e.g., CE 5 470 μf), we can estimate the input signal’s amplitude for second order harmonic distortion at the output. Note that Vin_mV_peak is measured in peak amplitude sinewave.

Harmonic distortion in percent 5 (Vin_mV_peak [ 1 Where re 5 re 5 (0.026 v/IC).

1                    2

] ) % 

1 (RE“||RE2/re)

For example, for Figure 8-31:

BT2 5 6 volts, R3 5 40.2KΩ, R4 5 20KΩ, RE” 5 3900Ω, RL 5 10KΩ,

RE2 5 1KΩ, G 5 –11.4

re 5 (0.026 v/IC) 5 (0.026 v)/(0.00033A) or re = 78.8Ω RE” || RE2 5 3900Ω || 1KΩ 5 796Ω

(RE” || RE2/re) 5 796/78.8 ~ 10

1 1 (RE” || RE2/re) 5 1 1 10 5 11

] ) % 

Second order harmonic distortion in percent 5

(Vin_mV_peak [ 1

1                    2

1 (RE“||RE2/re)


Second order harmonic distortion in percent 5 (Vin_mV_peak [ 1 ]2) % Second order harmonic distortion in percent 5 (Vin_mV_peak 3  1  ) %


So, if Vin 5 121 mV peakpeak (which is also 242 mV peak to peak), then: Second order harmonic distortion in percent 5 (121 3  1  ) % 5 1%

Vin’s peak to peak voltage for 1 percent second order distortion is then 2 3 121 mV peak to peak or 242 mV peak to peak. This 242 mV peak to peak input level assumes that the output has not clipped. With a gain G 5 –11.4, the output will be Vout 5 2.76 volts peak to peak (11.4 3 242 mV p-p).

Note that the 2nd order harmonic distortion formula is proportion to input level. That is doubling the signal level results in twice the distortion, or halving the input level gives half the distortion at the output.

A safer estimate to ensure that there is no clipping is to reduce Vin to the

0.5 percent level. This is just one half of 242 mV peak to peak input level for 121 mV peak to peak that results in the output.

Vout 5 1.36 volts peak to peak for 0.5 percent second harmonic distortion

Maximum Output Voltage Swing

We will now examine how to achieve maximum output swing in differently config- ured common emitter amplifiers.

Amplifier’s Emitter AC Grounded via CE

The DC collector voltage will determine output voltage swing. For the circuit in Figure 8-30, where capacitor CE is an AC short circuit, the maximum voltage swing will be from VE to BT2. For example, if VB is 2 volts, then VE 5 1.3 volts due to the 0.7-volt base – emitter voltage. If BT2 5 6 volts, then the peak to peak output swing will be from 11.3 volts to 16 volts, which will be (6 volts – 1.3 volts) peak to peak or

4.7 volts peak to peak. However, often the DC collector voltage is not centered to give maximum voltage swing without clipping prematurely on one-half of a sine wave cycle before the other half of the sine-wave. To maximize, you can set the DC collector voltage at the average of BT2 and VE, which is VC 5 (BT2 1 VE)/2.

For this example:

VC 5 (6 volts 1 1.3 volts)/2 or VC 5 3.65 volts for “symmetrical” clipping

Amplifier’s Emitter Partially AC Grounded via Series RE2 and CE2

In Figure 8-31, we have a little bit more going on because both emitter and collector terminals have AC output signals. Clipping occurs when either the collector voltage approaches the power supply voltage (e.g., BT2), or when the collector voltage matches the (same) voltage at the emitter. That is during the positive sine-wave cycle of the emitter and the negative cycle of the collector matches the voltage across the collector and emitter (e.g., VCE) is approximately 0 volt. Put in other words, the col- lector voltage equals (or matches) the emitter voltage.

For Figure 8-31 where RE2 is generally > re, the input signal range for “undis- torted” output can be approximated when RE” || RE2 >> re as follows: Turn off the input signal and measure DC voltages for VC and VE. Let’s call these DC voltages, VCdc and VEdc. The total collector voltage swinging downward toward the emitter’s voltage is VCdc1ac 5 VCdc 1 G Vinpeak and the total voltage at the emitter swinging up toward the collector voltage is VEdc1ac ~ VEdc 1 Vinpeak.

The input voltage that can be found by the collector voltage equals the emitter


VCdc1ac 5 VEdc1ac

VCdc 1 G Vinpeak 5 VEdc 1 Vinpeak

To find Vinpeak, this can be summarized as: Vinpeak 5 (VCdc – VEdc)/(1 – G)

A step-by-step calculation is now shown.

For example, BT2 5 6 volts, RL 5 10KΩ, R3 5 40.2KΩ, R4 5 20KΩ so

VB 5 BT2 [R4/(R3 1 R4)] 5 6 volts [20K/(40.2K 1 20K)] 5 6 volts [20K/60.2K]

VB ~ 6 volts [2/6] or VB ~ 2 volts. IC ~ VE/RE” 5 (VB – 0.7 volt)/3900Ω or IC ~ 1.3 v/3900Ω or IC 5 0.00033 A. This makes VCdc 5 6 volts – IC(RL) or:

VCdc 5 2.7 volts. VEdc 5 VB – 0.7 volts 5 2 volts – 0.7 volt 5 1.3 volts

VEdc 5 1.3 volts

We need to equate VCdc1ac 5 VEdc1ac to find the maximum peak voltage swing at clipping.

VCdc 1 G Vinpeak 5 VEdc 1 Vinpeak

VCdc 1 VEdc 5 G Vinpeak 1 Vinpeak 5 (1 – G) Vinpeak 5 (VCdc – VEdc) or

Vinpeak 5 (VCdc – VEdc)/(1 – G), to get a positive value in terms of peak input voltage:

Vinpeak 5 (VCdc – VEdc)/(1 – G) For this example:

VCdc 5 2.7 volts

VEdc 5 1.3 volts

G 5 –11.4

Vinpeak 5 (VCdc – VEdc)/(1 – G)

Vinpeak 5 (2.7 v – 1.3 v)/(1 – –11.4) 5 1.4 v/(12.4)

Vinpeak 5 0.1129-volt peak input The output peak voltage:

Voutpeak 5 Vinpeak 3 |G|

G 5 –11.4         |G| 5 11.4

Voutpeak 5 0.1129-volt peak 3 11.4

Voutpeak 5 1.287 volts peak

Voutpeak to peak 5 2.574 volts peak to peak

In practice both Vinpeak and Voutpeak will be slightly smaller to avoid clipping or distortion.

Finding an Optimum Bias Point for Maximum Output Swing with Just an Emitter Resistor

With a common emitter amplifier with just a series emitter resistor (e.g., RE) and where the series emitter resistor >> re 5 (0.026 v/IC), we can find an optimal bias voltage based on the collector load resistor, RL and series emitter resistor, RE. See Figure 8-35.

FIGURE 8-35 A common emitter amplifier with an emitter series resistor, RE.

To find an optimal biasing voltage, VB, given the values for RL and RE, we have: VB 5 BT2  1  1 0.7 volt

1 + RL


For example if we want to make a unity gain phase splitter amplifier where amplitudes of the AC signals are about equal via the collector and emitter terminals (Vout and Vout2), RL 5 RE.

NOTE: For a phase splitter amplifier circuit, output terminals Vout and Vout2 are generally loaded with equal resistance values of at least 10 x RL.

If BT2 5 12 volts, and RL 5 RE 5 1KΩ, then:

VB 5 12v   1   1 0.7 v 5 6 v 3 1 1 0.7 v


2  1 + 1K                                                        2


VB 5 3 v 1 0.7 v or VB 5 3.7 volts

This would mean VE 5 VB – 0.7 v or VE 5 3 volts.

If there is a 3-volt drop across RE, then there should be a 3-volt drop across RL since RE 5 RL, given that the emitter current 5 collector current for β >>1. That is VRL 5 VRE.

Thus, VC 5 BT2 – VRL 5 BT2 – VRE 5 12 v – 3 v or VC 5 9 volts.

This makes sense because at VE 5 3 volts, the maximum swing at the emitter is 3 volts ± 3 volts or 0 volt to 6 volts at the emitter. If there is 6 volts at the emitter for VE, then the collector voltage is VC 5 BT2 – VRE 5 12 v – 6 v, or VC 5 6 volts. This then satisfies the condition that maximum swing is when the collector and emitter voltages are equal.

In practice the collector-to-emitter voltage is rarely 0 volt but close, such as

0.2 volt. And this will reduce the maximum calculated swing by a slight amount. But this equation is fine for a first approximation.

VB 5 BT2  1  1 0.7 volt

1 + RL


With RL 5 RE 5 1KΩ, and BT2 5 12 volts, here are example resistor values for R3 5 39KΩ and R4 5 18KΩ. And we can confirm that VB is close to 3.7 volts.

VB 5 12 v [R4/(R3 1 R4)] 5 12 v [18K/(39K 1 18K)] 5 12 v [18K/57K]

VB 5 12 v [0.3158] or

VB 5 3.789 volts ~ 3.7 volts

Because RE 5 1KΩ and R3||R4 < 20RE  39KΩ || 18KΩ < 20KΩ 5 20RE, we can use a transistor of β ≥ 100 such as a 2N3904.


The amplifiers presented have limitations in terms of input amplitudes and output swing. Distortion can be a problem unless there is a series input resistor or a series emitter resistor that reduces gain and distortion. Be aware of the capacitors’ capaci- tance values at the input, at the output, and especially connected to the emitter (e.g., CE and CE2), which will determine the amplifier’s low-frequency performance.

Also, make sure electrolytic capacitors are biased correctly; otherwise, they will be reverse biased. Having electrolytic capacitors wired backwards results in leakage currents that will cause the expected DC bias points to shift up or down. You can use a voltmeter to confirm the correct polarity voltage across the electrolytic capacitors. This concludes Chapter 8 concerning simple one-transistor amplifiers. In Chapter 9 we will explore some linear integrated circuits such as operational ampli-

fiers and voltage regulators.

               CHAPTER 9

Analog Integrated

Circuits Including Amplifiers and Voltage


In Chapter 8, we examined some simple discrete amplifier circuits. We noted that these amplifiers include characteristics of input resistance, output resistance, voltage gain, output swing, and distortion. In particular, these amplifiers have limitations of output swing when connected to a low-resistance load at Vout, and distortion with large output voltage swings.

Many of the limitations of discrete amplifiers can be reduced by using integrated circuit (IC) operational amplifiers. Also, IC amplifiers take less space and can pro- vide lower current draw than discrete circuits. This chapter will explore some of the IC operational amplifier basic characteristics such as gain, drive capability, and power supply requirements.

In addition to IC amplifiers, we will look at voltage regulators. A voltage refer- ence (e.g., “super Zener diode”) with a power operational amplifier generally makes up a voltage regulator that provides a constant voltage even when the current load is large (e.g., > 100 mA).

Operational Amplifiers

Operational amplifiers, or “op amps,” historically were used to solve operational math operations such as adding, subtracting, integrating (e.g., averaging), and other opera- tions. When op amps are used with multiplying or nonlinear circuits, they can be used to find magnitudes of vectors (e.g., c 5 =a2 1 b2), or they can be used to find an inverse function such as a log function where the original function is an exponential function.


In general, an op amp, when used as an amplifier has the following characteristics:

•     A non-inverting input terminal.

•     An inverting input terminal.

•     An output terminal.

•     A positive voltage power supply terminal.

•     A negative voltage power supply terminal or a ground terminal.

•     A negative feedback element (e.g., resistor or capacitor) that is connected to the output terminal and the inverting input terminal.

•     When an op amp includes negative feedback, the voltage at the non-inverting input terminal is “equal” to the voltage at the inverting input terminal.

•     The non-inverting input terminal should be connected in such a way that it provides a DC path. Generally, the non-inverting input terminal should be connected to a voltage source via a straight connection or via a resistor or inductor. Otherwise, the op amp will drift up or down in voltage until the output voltage is clipped to one of the power supplies or ground voltages.

•     The inverting input terminal is connected via a resistor, inductor, and/or capacitor to the output terminal to form a negative feedback arrangement.

Operational amplifier U1 in Figure 9-1 shows an LM741 single package op amp with the non-inverting input (1) at pin 3, the inverting input (–) at pin 2, the output at pin 6 with the 1V supply at pin 7 and the –V supply at pin 4. The other pins 1 and 5 are generally not used unless the user wants to adjust the input offset voltage to zero. Ideally, the op amp will have perfectly matched input transistors that result in zero volts offset at its input. Zero volts offset means if you ground both inputs, the output voltage is 0 volts. However, in practice the input offset voltage can be within

± 10 mV DC, which means if you ground the two inputs, the output pin will not be zero volts. Generally, input offset voltages are not a problem except for instrumenta- tion and other measuring circuits where zero input offset voltage is required.

FIGURE 9-1 Op amp schematic symbols showing single, dual, and quad op amps in single packages of 8 pins (single and dual) and 14 pins (quad).

Analog Integrated Circuits      163

The LM741’s pin 8 is not connected. Note that U1’s power pins are 7 (1 v) and 4 (– v), with output pin 6, and input pins 2 (– input) and 3 (1 input). These pin assign- ments are standard for other single-package op amps such as TL081, TLC271, LF356, LF441, MC1456, and AD797.

U2A and U2B have standard pin outs for dual op amps in an 8-pin package. Here, pin 8 is the 1V supply pin and pin 4 is the –V supply. Note that the power pins are at the corners of the 8-pin package, and if the IC is installed backwards (e.g., the IC is turned around), the power pins 8 and 4 will receive a reverse voltage that will dam- age it. If this happens, you should replace the IC and ensure that the IC’s installation orientation is correct. For the first section, pins 2 and 3 are the inverting and non- inverting input pins, while pin 1 is the output. For the second section, pins 6 and 5 are the inverting and non-inverting input pins with output pin 7. Again, these pin outs are standard for dual op amps in an 8-pin package. Thus, dual 8-pin op amps such as the LM358, LM1458, LM4558, TL082, TLC272, LF353, LF412, ISL28208,

ISL28218, NJM4560, and AD712 will have the same pin outs.

For a 14-pin package, op amp sections U3A, U3B, U3C, and U3D show standard pin outs for the inverting and non-inverting input pins and corresponding output pins. Pin 4 is the 1V power pin, while pin 11 is the (negative) – V power pin. Again, for the most part these are standard power pins for a quad op amp in a 14-pin pack- age. Op amps with this pin out are LM324, TLC274, TL084, LM348. But there are exceptions such as the MC3301 quad op amp with 1V at pin 14, –V at pin 7. Therefore, if you are using a quad op amp, be sure to look up the data sheet.

NOTE: Again, if the IC is turned around, the power pins will be reversed and damage will be done to the quad op amp. You will have to replace the op amp and ensure that it is installed correctly.

Figure 9-2 shows dual inline packages (DIP) for through hole and surface mount 8-pin (SO-8, small outline 8-pin) and surface mount 14-pin (SO-14, small outline 14-pin) ICs.

FIGURE 9-2 8- and 14-pin DIP packages on the top row and 8- and 14-pin surface mount ICs on the bottom row.

We will now show a simple unity gain amplifier known as the voltage follower amplifier. See Figures 9-3 and 9-4. The voltage follower has close to zero ohms out- put resistance.

A voltage follower works on the negative feedback principle (Vout is connected to the inverting input). The signal voltage at the non-inverting input causes inverting input to have the same voltage. For example, in Figure 9-3 with U1, Vin1 is at pin 3 and at pin 2. But pin 6, the output pin (Vout1) is connected to pin 2 (inverting input). Thus, Vout1 5 Vin1. With dual power supplies such as ± 5 volts, the voltage follower has a gain of 1 for DC and AC signals. If the input signal is a DC signal (e.g., for instrumentation), the gain is 1. For example, if Vin1 5 11.56 volt DC, then Vout1 5 11.56 volt DC. Using bipolar (plus and minus) power supplies allows a lower-frequency response down to DC because a coupling capacitor is not required. These circuits will amplify AC signals as well. So why is a voltage follower useful? If the input voltage has little capability to drive larger currents, such as driving a head- phone or low-impedance load (e.g., < 1KΩ), using the voltage follower circuit allows for that. For example, if your input signal has an equivalent source resistance of 10KΩ, then only a small current (e.g., < 1 mA) will drive a 64Ω headphone, which will result in low volume. However, if you use an NJM4556 that can output up to 70 mA, driving the headphone will not be a problem.

Tekstvak: 165

FIGURE 9-3 Voltage follower amplifiers with dual supplies with decoupling capacitors (C1, C2, C3, and C4) connected directly to the power pins to avoid oscillations from the op amps.

The input resistances for each of the op amp voltage follower circuits are R1, R10, and R11, which can be changed to virtually any practical value. Thus, the input resistances of the voltage follower circuits as shown in Figure 9-3 are R1 5 100KΩ, R10 5 1MΩ, and R11 5 1MΩ. Input resistors R1, R10, and R11 can have different value resistors but we sometimes have to keep in mind the “tiny” input bias currents of the inverting and non-inverting input terminals. Input bias currents can cause a noticeable DC shift at the output when the product of the input bias and input resis- tors exceed 10 mV or more. If you choose an op amp that has an FET (Field Effect Transistor) input stage (e.g., TL082, LF353, TLC272), the input bias current is negli- gible and input resistors up to 10MΩ can be used without creating offset voltage problems.

Figure 9-4 shows a voltage follower circuit when using a single positive supply. When using a single power supply, generally, we need to create a half voltage sup-

ply to bias the input terminals and output terminal to half the supply voltage. The bias voltage, 1Vb 5 1VCC [R10/(R10 1 R11)]. Since R10 5 R11, 1Vb 5 1VCC [½].

Thus, pin 3 has the bias voltage 1Vb combined with the AC portion of the Vin14. For example, if Vin14 is a 1-volt peak to peak sine wave and 1VCC 5 9 volts, then 1Vb 5 4.5 volts, and pins 3, 2, and 1 will have a composite signal of 14.5 volts DC 1 1 volt peak to peak AC sinewave. The bias voltage of one half 1VCC provides for the largest AC signal swing at the output (e.g., pin 1 or 7). In Figure 9-4, one of the first things to look for in single supply op amp circuits is to measure referenced to ground that the two input terminals and the output terminal are at about one-half the supply voltage.

Figure 9-4 also shows the polarity connections for polarized electrolytic capaci- tors C40, C50, and C60. Note the positive terminal is connected to the input and output terminals that have positive DC voltages of 1VCC/2. Decoupling capacitor C30 is connected closely to pin 8, for example, within an inch to avoid parasitic oscillations from the op amp.

Also shown in Figure 9-4 is the second section of U2B that has its non-inverting input terminal pin 5 connected to U2A’s output pin 1. We can do this because pin 1 has the needed bias voltage 1Vb 5 1VCC/2, which “automatically” correctly DC biases pin 5 of U2B. Pin 7 then provides a second or extra output signal Vout15 via C60 essentially identical to Vout14. For example, if 1VCC 5 12 volts, then 1Vb 5 6 volts DC, and if U2 5 NJM4556, then we can drive two 64Ω earphones separately via Vout14 and Vout 15.

The AC input resistance with C40 being an AC short circuit, is R10 || R11, and with R10 5 R11 5 200KΩ, the AC input resistance 5 200KΩ || 200KΩ 5 100KΩ. Again, as a reminder R11 is connected to 1VCC, which serves as an AC ground. If you are only using a negative supply, see Figure 9-5.

Tekstvak: 167

FIGURE 9-4 A voltage follower amplifier with a single positive voltage supply to work with AC signals.

Tekstvak: 168

FIGURE 9-5 A voltage follower circuit with only a single negative power supply.

In this circuit, note that the 1V power pin 4 is connected to ground. We need to bias the input terminals to a voltage, –Vb such that –Vb 5 –VEE/2 . For example, if

–VEE 5 –5 volts, then –Vb 5 –2.5 volts DC.

Also, to avoid oscillations, decoupling capacitor C4 connects directly to pin 11. A common –VEE/2 voltage source can be made with a simple voltage divider

R13 and R14 and with a large value decoupling capacitor C40 as shown in Figure 9-5. C40 provides an AC short circuit to ground even at low frequencies such as 20 Hz, and it also provides an AC short circuit compared to the 100KΩ bias resistors R12 and R10. Any AC signal voltages coupled to R12 and R10 via input capacitors C70 and C90 at the input terminals of the op amps will be isolated from each other. This is because both lower connections of R12 and R10 go to C40, which is an AC short to ground. And because C40 is an AC short to ground, the AC input resistance to Vin16 is R12 (100KΩ), and the AC input resistance to Vin17 is R10 (100KΩ). Again, the (total) voltage at pins 3, 2, and 1 of U3A is –Vb plus the AC signal from Vin16. Likewise, the (total) voltage at pins 5, 6, and 7 of U3B is –Vb plus the AC signal from Vin17. Thus, the AC gain is 1. Note that –Vb can supply bias voltages via input bias resistors (e.g., 100KΩ) to the two other op amps U3C and U3D at pins 10 and 12.

In general, when making a voltage source from a resistive divider, you can scale the resistors from about 10 percent to 50 percent of the input resistors’ values, and scale the decoupling capacitor (e.g., C40) to at least ten times the capacitance of the input capacitors (e.g., C70 and C90).

Again, note the polarity of the electrolytic capacitors (e.g., C70, C80, C90, and C100) where the (–) terminals are connected to the input and output terminals of the op amps. Also the decoupling capacitor C40 has its (–) terminal connected to the voltage divider circuit R13 and R14 because R14 is connected to a negative voltage,


The voltage follower is a unity gain amplifier that can be characterized as an amplifier with almost infinite input resistance, Rin (not counting its input biasing resistor), and nearly zero output resistance, Ro. See Figure 9-6.

With plus and minus power supplies, the block diagram is valid for both AC and DC signals. The input resistance is just the input resistor RA that also serves to pro- vide a DC bias voltage (e.g., 0 volts DC) to the non-inverting input terminal of U2A. A voltage follower circuit has close to 0Ω output resistance. However, all op amps include output current limiting usually in the tens of milliamps to protect the op amp from destruction should the output terminal (e.g., pin 1) be accidentally shorted to ground.

For a single power supply where the bias voltage is provided by a voltage divider, see Figure 9-7 for the block diagram.

FIGURE 9-6  Model of a voltage follower with plus and minus supplies.

Because the circuit is AC coupled via input and output capacitors, C40 and C50, the block diagram refers only to AC signals for Vin and Vout. See Figure 9-7.

And as can be seen R11 is connected to the DC supply voltage 1VCC that is an AC short circuit to ground. Recall that by definition a pure DC supply voltage does not include any AC voltage. Thus, the AC input resistance is R10 || R11. When 1Vb 5 1VCC/2, R10 5 R11, so the input resistance 5 R10 || R10 5 (½) R10 or (½) R11. The AC output impedance provided by C50 is nearly an AC short circuit and is very small when compared to the subsequent load resistance at Vout. For example, if the load resistance, RL, at Vout to ground is 2kΩ and you need the amplifier to have a good low frequency response down to 20 Hz, a good rule of thumb is to set the cut-off frequency to about one-half to one-tenth of the lowest frequency desired. In this example, we can set the cut-off frequency between 2 Hz and 10 Hz. The low- frequency cut-off frequency 5 1/[2π(C50)RL] 5 1/[2π(10 μf)(2kΩ)] 5 7.96 Hz, which is within the range. However, electrolytic capacitors are relatively inexpensive and the cost of a 10 μf capacitor and a 33 μf capacitor is about the same. So, having

C50        33 μf will result in a low-frequency cut-off frequency of:

(7.96 Hz)/3.3 5 2.4 Hz. The 3.3 factor comes in because 33 μf is 3.3 times 10 μf.

FIGURE 9-7  Block diagram of a single supply voltage follower circuit.

We will now look into some very important specifications to keep in mind when working with op amps. The voltage follower configuration will be used as an example.

Maximum Safe Power Supply Voltage

Older op amps such as the LM741, TL081, LF351 and LM318 have a 36-volt maxi- mum across their V+ and V– terminals. Some of the newer single supply Texas Instruments LinCMOS™ amplifiers such as the TLC251or TLC271 have about a 16-volt maximum. Yet some of the newer op amps (e.g., LMV358 and NJM2100) only have a 6- or 7-volt supply limit.

See Table 9-1 on dual op amps.

TABLE 9-1 Various Op Amps with Maximum Voltage Across Their V+ and V– Supply Terminals

Op ampMax Supply Voltage ±Max Supply voltage + VPins V+ and V
LMV3582.5 volts5.0 volts8 and 4
TLC2728.0 volts16.0 volts8 and 4
LM35815.0 volts30.0 volts8 and 4
TL08216.0 volts32.0 volts8 and 4
ISL2821820.0 volts40.0 volts8 and 4
NE553221.0 volts42.0 volts8 and 4

Note: Max Supply voltage +V refers to a single supply

Although Table 9-1 shows the maximum supply voltages, to be on the safe side, the op amps are operated at ≤ 90 percent of maximum rated supply voltages. For example, the TLC272 would be operating with a 112-volt supply that is 4 volts below its maximum rating of 16 volts. Because not all volt meters (DVMs) are absolutely accurate, it’s always good to be on the safe side by running the op amps with a good safety margin below the maximum rated voltage.

Minimum Power Supply Voltage

Not all op amp will work correctly with low-voltage power supplies, such as a single

15-volt supply or at ± 3 volts.

For example, the LM1458 or TL082 op amp will not work properly or with very poor performance (e.g., very low output swing < 2 volts p-p) when used with a 12.5- volt and –2.5-volt supply. A new op amp such as the LMV358 will work at these voltages, but a higher supply voltage such as ± 5 volts will damage it. See Table 9-2 for the manufacturer’s minimum recommended supply voltages.

TABLE 9-2 Minimum Supply Voltages for Various Op Amps Via Data Sheets

Op ampMinimum Supply ± VoltageMinimum Single Supply Voltage
TLC2721.5 volts3.0 volts
LM3581.5 volts3.0 volts
ISL282181.5 volts3.0 volts
OPA21342.5 volts5.0 volts
OPA26044.5 volts9.0 volts
LM14585.0 volts10.0 volts

As an experiment using a voltage follower configuration (see Figure 9-6, U2A) Table 9-3 shows a summary of minimum supply voltages and their outcomes. The op amps are loaded into 10MΩ (essentially a no-load condition), and note that the results will be worse when the load resistance is 1KΩ.

TABLE 9-3 Experimental Results of Various Op Amps in Voltage Follower Configuration and with “No Load” at the Output

Op ampSupply ± VoltageOutput Voltage
MC1458± 2.5 volts0.6 volt peak to peak
TL082± 2.5 volts1.4 volts peak to peak
LM4558± 2.5 volts1.8 volts peak to peak
LF353± 2.5 volts2.5 volts peak to peak
OPA2134± 2.5 volts3.0 volts peak to peak
ISL28218± 2.5 volts3.0 volts peak to peak
TLC272± 2.5 volts3.5 volts peak to peak
OPA2604± 3.5 volts2.0 volts peak to peak
LT1632 (Rail to Rail in/out)*± 2.5 volts5.0 volts peak to peak

*  Note that the LT1632 op amp can still amplify properly over the entire supply range for the input and output. This type of op amp is called a “rail to rail” op amp for both input signal range and out- put voltage swing.

There are basically two types of rail to rail op amps.

•     An op amp that can swing the full supply voltage range, but a smaller input range that does not match the supply voltage(s). For example, for the ISL28218 op amp, if the power supply voltage is +12 volts, the output voltage swing is 0 volts to 12 volts, but the input voltage range would be 0 volts to about 10 volts.

•     An op amp that can swing the full supply voltage range that also has an input voltage range that matches the full power supply. For example, if the supply voltage is 5 volts, then the LT1632 has an input range from 0 volts to 5 volts while the output can swing from 0 volts to 5 volts also.

Caution on Providing Supply Voltages

We have to choose carefully the op amp to suit the power supply voltage or the op amp may be damaged due to overvoltage. But we must be aware of the minimum operating voltage for the op amps, or the op amp will not amplify the signal properly. If the incoming power supply has voltage fluctuations, then we need to ensure that there is no overvoltage to the op amps. One solution is to preprocess the raw

voltage into a regulated voltage using a voltage regulator integrated circuit.

For example, if you buy an unregulated wall charger DC supply that states it is a 12-volt DC 1 amp supply, usually it will provide a higher voltage (e.g., 17 volts) under a low-current load, while delivering the specified 12 volts DC at 1 amp. For safe operation, you can choose an op amp with a higher maximum voltage rating like an NE5532 that can operate up to 42 volts. Or you can use a lower-voltage op amp such as the TLC272 and regulate the “raw” voltage (e.g., a 12-volt to 17-volt range) by

using an 8-volt regulator chip such as an LM7808 to provide a constant 8.0 volts to the TLC272 op amp whose maximum supply voltage is 16 volts.

Alternatively, buy a regulated wall supply, which is usually a switching power supply.

To reiterate, when powering an op amp, it is best to use some type of voltage regulation. This way, if there are spikes in the AC power line, your op amps and other circuits (e.g., logic chips) will stay protected from damage.

Voltage regulator circuits will be discussed further at the end of this chapter.

Maximum Output Current

In general, most op amps have push pull class AB output stages that provide about 10 mA of positive and negative current output. With this type of output current capability, the op amp can drive resistive load ≥ 1000Ω with ≥ 10 volts output. But this 10-mA output current can be smaller if lower supply voltages such as a single 5-volt or 7-volt supply are used.

For example, a TLC272 amplifier with a 112 volts supply in a voltage follower configuration loaded into a 220Ω resistor outputs about 5.6 volts peak to peak, whereas with a 16-volt supply, the output (before distortion) drops to 2.25 volts peak to peak.

NOTE: Trying to pull more than the maximum current results in clipping or limiting to the output voltage.

If these types of op amps that have typically a 10-mA maximum output current are used to drive low-impedance headphones such as a 16Ω or 32Ω version, signal clipping will occur.

For example, a 32Ω load is driven directly with a TLC272 with 112-volt supply with a voltage follower circuit, as shown in Figure 9-8.

FIGURE 9-8  Voltage follower circuit.

The voltage follower circuit in Figure 9-8 having a 112 volt supply and a 32Ω resistor connected to the output produces a clipped (distorted) sine wave signal as shown in Figure 9-9.

FIGURE 9-9 Waveform output with TLC272 with 5 volts peak to peak input at 1000Hz, Vin1, and 2.69 volts peak to peak output at Vout1 as measured by the dashed horizontal lines/cursors.

There are amplifiers that can supply 20 mA to 30 mA such as LM833 or NE5532 dual op amps. These can drive lower-impedance loads such as 600Ω.

A high-current op amp such as the NJM4556 can output up to ± 70 mA. This allows driving low-impedance headphones (e.g., 16Ω to 64Ω stereo headphones), or even drive small loudspeakers with a 32Ω or 50Ω impedance. See Figure 9-10, where a 32Ω load is driven with an NJM4556 using 12-volt supply with a voltage follower circuit like the one shown in Figure 9-10. The NJM4556 peak output current into 32Ω is then the peak voltage, 2.5 volts peak/32Ω or 78 mA peak.

Note that 5 volts peak to peak is equal to 2.5 volts peak.

But there is a word of warning on some op amps with very limited output current such as the LM358 and LM324 that are commonly used in DIY (do-it-yourself) projects. These op amps have very poor output drive capability. Figure 9-11 shows an LM358 with a single 5-volt supply circuit driving a 10MΩ load (see Figure 9-8, where U2A 5 LM358 and VCC 5 5 volts, and R3 5 10MΩ or 2000Ω). With a 3000 Hz 0.5-volt peak to peak signal at Vin, see Figure 9-11 for the output waveform.

FIGURE 9-10 Waveform from NJM4556 delivers a clean sine wave with 5 volts peak to peak into 32Ω. Note that 5 volts peak to peak is the same as 2.5 volts peak.

FIGURE 9-11 An LM358 voltage follower circuit’s output signal into a 10MΩ resistor with a clean 500-mV peak to peak sine wave.

However, if the LM358 voltage follower drives a 2000Ω (for R3 in Figure 9-8) instead of the previous 10MΩ, the waveform distorts quite badly. See Figure 9-12.

FIGURE 9-12 The LM358 voltage follower’s distorted output waveform (~ 500 mV peak to peak) when driving a 2000Ω resistor (R3). Note the discontinuities near the “zero” crossing of the sine wave. This is known as crossover distortion.

One way to reduce or eliminate the crossover distortion in an LM358 or LM324 is to connect a pull-up or pull-down resistor (e.g., 1KΩ or 1.5KΩ) from the output terminal to either the 1V or –V terminal or ground terminal. See Figures 9-13 and 9-14.

FIGURE 9-13 A pull-down resistor R4′ is connected from the output pin 1 to the –V or ground terminal pin 4 of the LM358 op amp to reduce or eliminate crossover distortion. Alternatively, you can use a pull-up resistor tied to pin 8 or +VCC as shown in Figure


FIGURE 9-13 A pull-down resistor R4’ is connected from the output pin 1 to the –V or ground terminal pin 4 of the LM358 op amp to reduce or eliminate crossover distortion Alternatively, you can use a pull-up resistor tied to pin 8 or 1 VCC as shown in Figure 9-14. 

FIGURE 9-14 A pull-up resistor, R4″, is connected to the LM358’s output pin 1 and pin 8 (+VCC) to reduce or eliminate crossover distortion.

The modifications via R4′ and R4″ in Figures 9-13 and 9-14 result in a cleaner looking sine wave as shown in Figure 9-15.

FIGURE 9-15  LM358 with a pull-down or a pull-up resistor that eliminates crossover distortion on the sine wave. The cursors with dashed lines across are also displayed for 500 mV peak to peak.

Output Voltage Range

When we supply an op amp with ± supply voltages such as ± 5 volts, we would expect that the output can swing approximately ± 5 volts. In practice, the op amp generally has a more restricted range. For example, in a TLC272 op amp, there is about a 1-volt loss from the V1 supply, but the op amp can swing almost to V–. If we use ± 5 volts, the output range is about (5 volts – 1 volt) to –5 volts, or 14 volts to –5 volts. Of course, the output range will further reduce when loaded into a resistor that draws at least 1 mA of current.

Other op amps can have more loss such as the LM 741 that loses about 2 volts on both V+ and V– supplies. For example, with a ± 9-volt power supply, the output swing is about 17 volt to –7 volts.

Special op amps that are called “rail to rail” usually provide just that—output swings that match the supply voltages. For example, an ISL28208 will provide nearly the same output swing voltages as the power supply within about 100 mV. However, because the ISL28218 op amp does not have rail to rail performance at the input terminals, when it is configured as a voltage follower the output swing will be restricted to the input range, which loses about a volt or two. For example, in Table 9-3 the ISL28218 has a swing of 3 volts with a 5-volt power supply when it is config- ured as a voltage follower. However, we will see later that we can take full advantage of the rail to rail output voltage range even if the input voltage range is restricted by using the amplifier as an inverting gain amplifier or as a non-inverting gain ampli- fier where the gain is generally ≥ 2.

Input Signal Range

Even though an op amp can swing rail to rail, that does not ensure that the input range is rail to rail. That is, if the power supply voltages are ± 5 volts, most op amps have a common mode input range that is less than or within the power supply volt- ages. For example, the CA3130 (Renesas) op amp can swing rail to rail at its output, but the input has a range of V– to V+ –2 volts. For a 5-volt supply where V– 5 ground or 0 volts and V+ 5 15 volts, the input range is 0 volts to V+ –2 or 0 volts to 3 volts. So, in a voltage follower circuit even if the input signal is from 0 to 5 volts, the output will be 0 to 3 volts.

See Table 9-4 for some op amps with various output swing and input ranges for a single positive voltage supply, 1V, that is referenced to ground.

TABLE 9-4 Some Op Amps’ Input and Output Voltage Ranges for a Single Supply

Op AmpInput Voltage RangeOutput Voltage Range
LM358 or LM324–0.2 v to +V – 1.5 v+0.6 v to +V – 1.5 v
LM1458 or LM741+3.0 v to +V – 0.5 v+1.7 v to +V – 1.0 v
NE5532 or NE5534+2.0 v to +V – 3.0 v+1.5 v to +V – 1.5 v
TLC2720 v to +V – 1.0 v0 v to +V – 1.0 v
ISL282180 v to +V – 2.0 v0 v to +V
LT16320 v to +V0 v to +V

Let’s take examples where 1V 5 6.0 volts or four 1.5-volt cells connected in series.

See Table 9-5.

TABLE 9-5 Examples of Voltage Ranges Where the Single Supply Voltage Is + 6.0 Volts

Op AmpInput Voltage RangeOutput Voltage Range
LM358 or LM324–0.2 v to +4.5 v+0.6 v to +4.5 v
LM1458 or LM741+3.0 v to +5.5 v+1.7 v to +5.0 v
NE5532 or NE5534+2.0 v to +3.0 v+1.5 v to +4.5 v
TLC2720 v to + 5.0 v0 v to + 5.0 v
ISL282180 v to + 4.0 v0 to + 6.0 v
LT16320 v to + 6.0 v0 v to + 6.0 v

A main takeaway from Tables 9-4 and 9-5 is that there are only certain types of op amps that will work well with a single 5 volt or 6 volt supply such as the LM358, ISL28218, TLC272, and LT1632. The traditional op amps used with higher supply voltages such as the LM741, LM1458, and NE5532 do not generally work at low voltages.

Also, if you notice carefully, op amps that state single supply capability have an input voltage range of at least down to ground, 0 volts, or the –V supply pin of the op amps. See the input ranges of the LM358, ISL28218, TLC272, and LT1632. The abil- ity to “sense” input signal near ground or the –V supply of the op amp can be very handy. See Figure 9-16, where we can make a constant current circuit out of an LM358.

Q1’s collector provides a constant DC current to light-emitting diode LED1 by regulating the emitter voltage, VE to 0.1 volt DC. By having emitter voltage stable at

0.1 volt the emitter current, IE, then is 0.1 volt/5Ω or IE 5 20 mA. Since IE 5 IC, the collector current, IC 5 20 mA. Since the collector is supplying current to the LED, the LED current is also 20 mA.

So, the question is how do we get a regulated 0.1 volt at the emitter? The answer is by negative feedback, which is done in the following manner:

FIGURE 9-16 A feedback current source circuit with nearly 0 volt sensing at the input terminals of the LM358 op amp at pins 3 and 2.

Via voltage divider R1 and R2 with the 5-volt supply voltage, we have 0.1 volt at the (1) input of the op amp U1A. Because there is negative feedback, the (1) input terminal’s voltage at pin 3 forces the (–) input terminal’s voltage at pin 2 to be the same voltage. Thus, pin 2 has 0.1 volt. If we follow where pin 2 is connected to, pin 2 is connected to Q1’s emitter. To force the 0.1 volt at the emitter, the base voltage will have to be high enough to cause conduction in the transistor via forward biasing the base-emitter junction. The base-emitter voltage is VBE ~ 0.7 volt. So, the base volt- age referenced to ground has to be VB 5 VBE 1 VE or VB 5 0.7 volt 1 0.1 volt. Thus, VB ~ 0.8 volt. By observing that if we make the approximation that the base current is small, then the divide-by-two voltage divider via R3 and R4 will cause the drive voltage into R3 to be twice that of VB. Therefore, the output pin 1 has to be twice VB or 2 3 0.8 volt. Thus, we get 1.6 volts at pin 1, which is within a “comfort- able voltage output range” for the op amp. Because this is a negative feedback circuit, if the transistor’s VBE changes with temperature to a lower or higher voltage than

0.7 volts, the output pin 1 will readjust automatically to ensure that there is always

0.1  volt at Q1’s emitter terminal.

NOTE: The voltage divider via R1 and R2 provides the 0.1 volt via 5 volts [R2/(R1 + R2)] or 5 volts [6800/(330,000 + 6800)] =

5 volts [0.02019] = 0.10 volt.

Non-Inverting Gain Amplifiers

Previously, we discussed about voltage follower circuits that have a voltage gain of 1. They have very high input resistance set by the input bias resistor to the (1) input, and a very low output resistance that is capable of driving loads typically in the 2KΩ range or lower. The voltage follower preserves the signal’s phase from input to out- put, whereas in Chapter 8, we saw that a common emitter amplifier’s output inverts the phase of the input signal.

Op amps can be configured to have the advantages of the voltage follower while providing voltage gains ≥ 1. See Figure 9-17.

FIGURE 9-17  An example non-inverting amplifier with plus and minus power supplies.

The amplifier shown in Figure 9-17 is directly coupled to the input signal source, Vin1, and its output is directly coupled to a load resistor, R4. Load resistor R4 is just shown as an illustration and can be omitted. In essence this amplifier will amplify not only AC signals, but also DC signals.

Note that an input bias resistor, R3, is required so that even if the input signal Vin1 is disconnected, the op amp will be properly biased at the input terminal. Although R3 is shown as 100KΩ, it can be almost any value within reason. That is, it does not make sense to have R3 5 1Ω because that will load down the input signal source, Vin1, and drain excessive current, or worse, cause the input signal source to be damaged. Typical input bias resistances for R3 are in the range of 100Ω to 10KΩ when using op amps with bipolar input stages such as the NE5532 or LM741, etc. For op amps that have field effect transistor (FET, MOSFET, or JFET) input stages such as the TLC272, TL082, and OPA2134, the input bias resistor R3 can be as high as 10MΩ.

If in doubt, you can find a variety of field effect transistor input op amps by a web search or by searching your op amp at or and specifying an input bias current < 2 nA (< 0.002 μA or < 2000 pA) at the 25-degree Celsius/centigrade rating. For example, the TL082 has a maximum input bias cur- rent 200 pA rating at 25 degrees C.

A general rule of thumb is if the input bias current multiplied by the input resis- tance (R3) or the feedback resistor (R2) is less than 10 mV, then you are OK. For example, if we used an LM1458 op amp, with a worst case 500 nA input bias current, then we want 500 nA 3 R3 < 10 mV or R3 <10 mV/500 nA, R3 < 20KΩ. This is just a general rule of thumb, and if you are working in measuring or instrumentation circuits, then it may be best to choose a low offset voltage (Vos) field effect transistor op amp such as the OPA2134 or LF412.

For a non-inverting gain amplifier, the gain is:

Av 5 Vout1/Vin1 5 (R1 1 R2)/R1

For example, in the circuit, R1 5 2KΩ and R2 5 18KΩ:

Av 5 (R1 1 R2)/R1 5 (2KΩ 1 18KΩ)/2KΩ 5 20KΩ/2KΩ 5 10

Av 5 10

For non-inverting gain op amp circuits, the gain is Av 5 (R1 1 R2)/R1 and the output resistance at Vout1  0Ω. That is, we normally do not have to worry about having the gain reduced if we load the output of the op amp with any range of resis- tance values as long as the op amp’s output stage can deliver sufficient current. But be aware that normally the output current limit is in the ± 10 mA range.

Now let’s intuitively understand how gain equation:

Av 5 Vout1/Vin1 5 (R1 1 R2)/R1 is derived. (See Figure 9-17.)

Recall the voltage at the (1) input terminal and the (–) input terminal is the same. In a negative feedback system, the (1) input terminal (e.g., U2A pin 3) deter- mines the voltages at the output and (–) input terminal.

As a reminder note that you have to ensure that the (1) input always has a DC path to a voltage source. If you leave the (1) input unconnected or coupled to a capacitor, your circuit will not work. Now let’s go back to how this circuit works.

We see that R1 and R2 form a voltage divider for the output voltage, Vout1. That is: Vout1[R1/(R1 1 R2)] 5 voltage at the (–) input terminal (e.g., U2A pin 2 in Figure 9-17). But we know that the (–) input terminal’s voltage 5 (1) input termi- nal’s volts, which is Vin1.

Therefore, Vout1[R1/(R1 1 R2)] 5 Vin1, which leads to:

[Vout1/Vin1][R1/(R1 1 R2)] 5 1

and if we divide by [R1/(R1 1 R2)] on both sides of the equation:

[Vout1/Vin1] 5 1/[R1/(R1 1 R2)] 5 (R1 1 R2)/R1

Vout1/Vin1 5 (R1 1 R2)/R1

This confirms the previously stated gain equation for Av for a non-inverting gain amplifier. What’s happening inside the op amp is that it is providing a voltage at Vout1 such that when the voltage is attenuated via R1 and R2, the voltage at the (–) input terminal matches the voltage at the (1) input terminal, which is Vin1. This is how a negative feedback system works. So even if you try to load down the op amp by lowering the resistance of R4, the op amp will compensate to deliver a voltage at Vout1 that ensures that the voltage at the (–) input terminal matches the voltage at the (1) input terminal.

But also note that the op amp has output current limitations, and we have to be mindful how we set the gain of the amplifier while working within the output cur- rent capability of the op amp. The feedback resistors R1 and R2 also load the op amp at Vout1. For example, we will get into trouble if we set R1 5 2Ω and R2 5 18Ω. The op amp’s output terminal will see the (R1 1 R2) as an “extra” load resistor, which will be 2Ω 1 18Ω or 20Ω. If the op amp has a 10-mA maximum output current, then the maximum output voltage (e.g., at U2A pin1) will be ± 20Ω 3 10 mA or ± 200 mV, which is a very limited output swing.

Often it is better to protect your input terminals from external DC bias voltages riding on top of AC signals. There are some devices that will provide an output sig- nal with a DC bias voltage, which will cause a directly coupled amplifier such as in Figure 9-17 to overload. For example, if 1VCC 5 16 volts and –VEE 5 –6 volts and Vin1 5 (1 volt DC 1 VAC_signal), and Av 5 10, then:

Vout1 5 10 3 Vin1

Vout1 5 10 3 (1 volt DC 1 VAC_signal) or

Vout1 5 10 volts DC 1 10 VAC_signal

However, since the power supplies are ± 6 volts, there is no way that the op amp U2A can provide 10 volts DC at Vout1. Therefore, the amplifier will be clipped to the positive rail at Vout1 and there will be no AC voltage at Vout1. We can fix this “prob- lem” by AC coupling via an input capacitor. See Figure 9-18.

By using an input coupling capacitor, C2, any DC bias voltage riding on top of the AC signal from Vin1 is eliminated at the (1) input terminal (e.g., U2A pin 3). Thus, the DC voltage at pin 3 is 0 volts, which leads pin 2, the (–) input, to have also 0 volts DC, and this further leads to the DC voltage at pin 1 is also 0 volts. Capacitor C1 at 1μf is used to provide an AC short circuit to ground such that the AC gain is still (R1 1 R2)/R1 or 10 in this example with R1 5 20KΩ and R2 5 180KΩ. If you notice carefully, the resistors R1 and R2 have been scaled up tenfold from Figure 9-17. Why? The reason is if we want the same low-frequency response with R1 5 2KΩ and R2 5 18KΩ, C1 would have to be 10 μf. In practice, it is harder to find a 10 μf non- polarized capacitor when compared to obtaining a 1 μf capacitor.

FIGURE 9-18 Using an input capacitor, C2, to block any DC voltage from Vin1 while passing AC signals.

Capacitor C1 also is used to prevent amplifying the op amp’s internal offset volt- age, Vos, which is usually in the 5 mV to 10 mV range for general purpose amplifiers. If C1 were replaced with a 0Ω resistor, then we have a DC gain of 10, which will result in a DC voltage at the output of 10 3 Vos. For example, if Vos 5 ± 10 mV, then the output DC voltage can be ± 10 3 10 mV or ± 100 mV DC. With C1, the offset voltage is just voltage “followed” and the DC gain with C1 is 1. So, with C1 installed, Vout1’s DC voltage is just Vos or typically ≤ 10 mV DC.

Note that the built-in offset voltage in the op amp can be a positive or negative DC voltage. An ideal op amp will have Vos 5 0 volt. You can buy precision op amps with Vos ≤ 1.0 mV such as the LF412(note the LF412A has a tighter Vos specifica- tion ≤ 1.0 mV than the LF412 without the “A” designation, which has Vos ≤ 3.0 mV). Figure 9-18 also shows that you can AC couple the output signal via C5 with Vout1 or have the output signal DC coupled via Vout1″. An AC coupled output is useful for blocking any DC voltage that may go into the output terminal of U2A. This is important in some cases where the output terminal of an amplifier is driving another circuit that has a DC bias voltage. If you use the Vout1″ terminal, the DC bias voltage from the other circuit will be “shorted” DC wise to ground or zero volts that may cause this other circuit to have the wrong DC bias and not operate cor- rectly. Thus, to play it safe, use a DC blocking output capacitor, C5, and also remove

the load resistor R4.

If you are using a single power supply non-inverting amplifier, see Figure 9-19.

FIGURE 9-19  A single power supply non-inverting gain amplifier.

The AC gain of this amplifier is again Av 5 (R1 1 R2)/R1. To maximize on volt- age swing, the DC bias voltage at the (1) input is set at 1VCC/2. For example, if 1VCC 5 9 volts, the DC bias voltage is set to 4.5 volts at pin 3 of U2A via voltage divider R3 and R6, where R3 5 R6. The AC input resistance is R3 || R6, which is 100KΩ || 100KΩ or 50KΩ. Note that resistor R6 is connected to 1VCC, which is an AC ground.

Because we are using a single supply, we can use polarized electrolytic capacitors (C1, C2, and C3) that allow for higher capacitance values at low cost.

To analyze this circuit DC-wise, see Figure 9-20 where we remove the capacitors C1, C2, and C3. Since these capacitors block DC currents when charged to their final value, they are open circuits in terms of DC analysis.

By removing the capacitors, a simplified circuit is shown in Figure 9-20 on the right. On the left side circuit, since R1 and R4 are disconnected on one end, they can be removed, which is shown on the right side circuit.

Here we see only a voltage divider circuit with R3 and R6, along with R2, the feedback resistor. However, because the op amp’s (–) input terminal has a high input resistance, there is no current flowing through R2. This implies there is no voltage across R2 since the voltage across R2 is equal to VR2 5 I2 3 R2, where I2 5 current flowing into the (–) input terminal. Since I2 5 0 amp flowing into the (–) input:

VR2 5 0 amp 3 R2 5 0 volt

Therefore, we can simplify further this right side circuit. See Figure 9-21.

Tekstvak: 187

FIGURE 9-20 Removing C1, C2, and C3 simplifies the circuit for DC analysis.

FIGURE 9-21 With zero current I2 flowing through feedback resistor R2, the circuit reduces to a simple voltage follower amplifier for DC analysis.

Because there is 0 volts across R2, this means the voltage at output pin 1 is the same voltage at pin 2, the (–) input. This then means we can replace R2 with 0Ω or a wire since we will get the same result, the voltage at pin 1 is the same as the voltage at pin 2. Thus, the DC voltage at pin1 is just 1VCC/2 since the voltage at the (1) input pin 3 is the same as the voltage at (–) input pin 2 when negative feedback is applied.

Figure 9-22 shows a model of a non-inverting gain amplifier for AC signals.

FIGURE 9-22  A generalized model for the non-inverting amplifiers, with G = Av.

From Figure 9-22 we have: Ro ~ 0Ω

NOTE: Keep in mind that op amps generally deliver about ± 10 mA into a load.

Ri 5 R3 in Figures 9-17 and 9-18, Ri 5 R3 || R6 in Figures 9-19 and 9-20.

Typically, Ri ≥ 10KΩ.

G 5 Av 5 [(R1 1 R2)/R1] for Figures 9-17, 9-18, 9-19, and 9-20.

Also, be sure to choose resistance values for the feedback network R1 and R2

such that it does not load the op amp’s output current excessively. Determine the maximum signal output and divide by (R1 1 R2) to determine the current draw from the op amp. Generally, this current draw should be less than 50 percent of the op amp’s maximum output current.

For example, in Figure 9-7, if the maximum current output is 10 mA and you are using ± 12 supplies, for Av 5 5 you can use R2 5 4020Ω and R1 5 1000Ω. Note Vsupply 5 12 volts.

Worst case, the load current due to the feedback network will be:

Vsupply/(R1 1 R2)

For this example, it will be 12 volts/5020Ω. This translates to about 2.4 mA, which is less than the 10-mA output limit. So, this is OK.

On the other hand if you choose R2 5 120Ω and R1 5 30Ω, the current draw will “want to be” 12 volts/(120Ω 1 30Ω) 5 80 mA.

This will cause the amplifier to clip at the output, and the waveform will clip at Voutput max ≤ (R1 1 R2) 3 10 mA or ≤ 150Ω 3 10 mA or ≤ 1.5 volts peak.

If a non-inverting gain amplifier circuit is generally set for a gain or Av ≥ 2 with

an op amp that has rail to rail output voltage but does not have rail to rail perfor-

mance at the input terminal, then this circuit can deliver rail to rail at the output while keeping within the input voltage range for proper operation. The reason is that by having a gain of ≥ 2, the input voltage range no longer has to go from rail to rail, but instead ≤ 50 percent of the output voltage range. For example, suppose a single supply amplifier shown in Figure 9-17 has an input range of 1VCC – 1.2 volts to

–VEE, and the power supplies are 1VCC 5 12.5 volts and –2.5 volts. This means the input terminal must be in the range of –2.5 volts to (12.5 volts – 1.2 volts), which is between –2.5 volts to 11.3 volts. Now suppose the gain is 12 by having R1 5 R2 5 10KΩ. At the zero-signal condition, the output voltage is 0 volts.

If the input signal is between –1.25 volts and 11.25 volts (e.g., a sine wave input signal whose peak amplitude is 1.25 volts), then the output will be twice that due to setting Av 5 2. Therefore, the output voltage will swing from –2.5 volts to 12.5 volts. And note that the input voltage signal range of –1.25 volts and 11.25 volts is within the input range of the op amp at –2.5 volts to 11.3 volts.

Troubleshooting Non-Inverting Op Amp Circuits via an Oscilloscope, Signal Generator, and DVM

Here are some tips on basic debugging techniques:

1.            Make sure that the power supply terminals are connected correctly and use a voltmeter to confirm proper power supply voltages at the power pins. Be sure to have the op amp’s data or spec sheet handy.

2.            Make sure there is at least a ≥ 0.1 μf decoupling capacitor wired or soldered closely to each power pin (e.g., plus supply pin 8 and negative supply pin 4 of an LM1458) and ground. For example, see the 1 μf power supply decoupling capacitors C4 and C3 in Figure 9-17. For a single supply circuit, you need only one decoupling capacitor since the other power pin is usually grounded. See Figure 9-19 where positive supply pin 8 of U2A is decoupled with 1 μf capacitor C4 to ground, while its negative power pin 4 is connected to ground. Without power supply decoupling capacitors, op amps can oscillate or give noisy output signals on their own.

3.            Confirm that the input DC voltages are within the range as specified by the data sheet. For example, an LM358 at 15 volts power supply has an input voltage range from about 0 volts to 3.5 volts. Use a voltmeter to confirm that the non-inverting input terminal (e.g., pin 3 or 5 in an LM358) is within the input voltage range.

4.            Measure the voltage with a DVM across the non-inverting input pin and the inverting input pin of the op amp. Confirm that the voltage is ~ 0 volt. If the voltage is not zero such as 0.5 volts DC, then the op amp circuit is not operating properly.

5.            For op amp circuits with positive and negative supplies, verify that the DC output voltage is close to zero. Verify that the non-inverting input terminal is connected via a wire or resistor (or a coil) to a DC potential such as ground or a bias voltage. With an ohm meter (DVM) and with the power supplies turned off, confirm that there is some resistance from the non-inverting input terminal to ground to a DC voltage line. For example, in Figure 9-18, the non- inverting input pin 3 is connected to ground via a 100K ohm resistor (R3) to establish a 0-volt DC bias.

6.            For op amp circuits with positive and negative supplies, if you are getting unexpected DC voltages at the output pin, then with the input signal source disconnected, try temporarily connecting the non-inverting input terminal to ground via a resistor (e.g., 10KΩ or lower) and re-measure the DC output voltage. If it is now correct, look for a bad bias resistor or bad connection in the non-inverting input terminal area. If connecting the non-inverting input terminal to ground does not help, then temporarily connect a wire or low- value resistor (e.g., 100Ω) from the output to inverting input terminal to form a voltage follower. If you now get a DC voltage ~ 0 volts, then recheck the feedback resistor network for the correct values. But it may be easier to just use a DVM and measure the resistance of the feedback network (e.g., R1 and R2 in Figure 9-7) with the power turned off.

7.            If after #1 to #6, you are not getting anywhere, replace the op amp, but make sure the original op amp could operate at the power supply voltages. For example, if the power supply provides ± 12 volts DC and the circuit does not seem to work, verify that the op amp is not a lower voltage version such as an

LMV358 (e.g., 15.5 volts maximum) or a TLC272 where a supply voltage of more than ± 8 volts will cause damage.

8.            If the DC bias conditions are correct, but the AC signals are distorted or do not have correct amplitudes given the gain setting [(R1 1 R2)/R1], then first check the resistance values of the feedback resistors. Also make sure the feedback network resistance value of (R1 1 R2) ≥ 500Ω is not low enough to cause current limiting at the output that will cause clipped waveforms.

9.            If the low-frequency (e.g., 20 Hz) response is poor—such as less than 50 percent of amplitude when compared to the signal amplitude at a mid-band frequency (e.g., 1000 Hz)—check the input and output coupling capacitors. If they measure correctly, check the frequency response at the non-inverting input pin (e.g., pin 3 or 5 in a dual op amp). If that’s OK, check the frequency response at the output pin of the op amp such as pin 1 or 7 of a dual op amp. If there is a loss in low-frequency response, see if there is an AC coupling capacitor in the feedback network, such as C1 in Figure 9-18 or C1 in Figure 9-19. If you suspect there is a problem like the capacitance is not high enough, parallel a larger capacitance across C1, such as 2.2 μf capacitor. If the capacitor is ceramic, check the maximum working voltage. For example, if you use a 1-μf, 6.3-volt capacitor and the DC voltage across it is 15 volts, chances are that the true capacitance will be half of that, or 0.5 μf, which will cause a poorer low-frequency response.

10.            Also, if electrolytic capacitors are used, measure with a volt meter (DVM) that the DC voltage across the capacitor matches the polarity markings of the capacitor. If you measure a reverse bias voltage across an electrolytic capacitor, turn off the power immediately, replace the capacitor and install a new capacitor with the correct polarity.

11.            Another troubleshooting technique is to confirm that the feedback decoupling capacitor is an AC short circuit. Apply a sine wave signal between 200 Hz and 500 Hz at the input and measure the AC voltage across the C1/R1 junction (e.g., of Figure 9-19). There should be a very small AC voltage there when compared to the AC signal at the non-inverting input pin (e.g., pin 3 or 5 in a dual op amp).

12.        Also check if the circuit was for a dual op amp but a single op amp was installed instead, or vice versa. Both single and dual op amps have 8 pins total and some pins are common, such as the –V supply pin 4, non-inverting input pin 3, and inverting input pin 2; however, the other pins 1, 5, 6, 7, and 8 are not the same for a single op amp and dual op amp.

This concludes an introduction to non-inverting gain op amp circuits. Next, let’s take a look at inverting gain amplifiers.

Inverting Gain Amplifiers

We now turn to inverting gain op amp circuits where they produce an inverted waveform at the output when compared to the input signal. See Figure 9-23.

FIGURE 9-23  An inverting gain op amp circuit with plus and minus power supplies.

Tekstvak: Side Note: How Does the Inverting Gain Amplifier Work, Really?
An inverting gain op amp circuit also still relies on the voltage applied to the (+) input terminal that defines how the output signal will react to match the (–) input’s voltage to be the same as the (+) input’s voltage. For a simple example, let R1 = R2 in Figure 9-23. If Vin1 = +1 volt DC, what voltage must Vout1 be such that at the junction of R1 and R2 that is connected to the (–) input matches the voltage at the (+) input terminal? We see that the (+) input termi- nal is grounded or at 0 volts. R1 and R2 form a voltage summing circuit for Vin1 and Vout1. If the resistors R1 and R2 are equal in value (e.g., R1 = R2 = 2KΩ) and we have + 1 volt DC into R1 via Vin1, it makes sense that the output voltage at Vout1 should be – 1 volt DC. This way at the junction of R1 and R2 we will get 0 volts. The output terminal (e.g., pin 1) is adjusting its voltage such that when summed via R2, the voltage at pin 2, the (–) input terminal, is zero volts. This output voltage adjustment via R2 provides the “virtual” short circuit voltage across the (–) input and (+) input terminals, with a voltage applied to input resistor R1.

The gain Av 5 Vout1/Vin1 5 –R2/R1. So, if we keep the same feedback resistors as in the circuit of Figure 9-17, that is, R1 5 2KΩ and R2 5 18KΩ, we find that Av 5 –18K/2K or Av 5 –9. Note the magnitude of the gain for the inverting ampli- fier: –R2/R1 is different from the non-inverting gain amplifier’s formula of

(R1 1 R2)/R1 5 1 1 R2/R1. The difference between the magnitudes of the two gains of non-inverting amplifier and inverting amplifier is just 1. For example, the non-inverting gain amplifier will have a gain of 10 with R1 5 20KΩ and R2 5 180KΩ, while the inverting gain amplifier will have a gain of –9.

The input resistance is just R1. So why is this? The non-inverting input is con- nected to ground; its voltage is 0 volts or Vi (1) 5 0 volts. Since in a negative feed- back amplifier the voltage at the inverting input terminal is the “same” as the voltage in the non-inverting input terminal, we can deduce that the voltage at the inverting input terminal Vi (–) is also 0 volts. Thus, R1 on the right side connection to pin 2 of U1A is “virtually grounded.” Therefore, referenced to ground, the input resistance is R1. In other words, if we look at an equivalent resistor referenced to ground from the inverting input node Vi (–), its resistance is Rsc, where Rsc is a short circuit to ground, equivalently. Rsc is a very low equivalent resistor because there is a large magnitude open loop gain inside the op amp. It is very similar to the feedback resis- tor in a one-transistor amplifier from Chapter 8 where we found that with a high- gain amplifier the input resistance looking into the base with the collector-to-base resistor is very low.

Because of R2, the feedback resistor, Rsc 5 R2/(1 1 a), where “a” is the open

loop gain. For example, if we have an NE5532 amplifier whose open loop gain is

~ 20,000 at 2 kHz, then the Rsc 5 R2/(1 1 20,000) ~ R2/20,000. Thus, if R2 5 180KΩ, Rsc ~ 180KΩ/20,000 or Rsc ~ 9Ω.

Since R1 5 20KΩ and Rsc 5 9Ω << 20KΩ, the 9Ω Rsc resistor equivalently shorts to ground the right side lead of R1 at Vi (–).

In terms of the output resistance at Vout1, the op amp with the negative feedback resistor R2 has an output resistance close to zero ohms. The amplifier in Figure 9-23 amplifies DC and AC voltages at Vin1. For example, if Vin1 5 1.0 v DC, then Vout1 5 –(R2/R1) 1 1 volt DC. For example, if R2 5 180KΩ and R1 5 20KΩ, the output voltage will be: Vout1 5 –(180K/20K) 3 1 volt DC or Vout 5 –9 volts DC. Note that the supply voltage should be at least a few volts higher than 9 volts, such as

± 12-volt supplies for 1VCC and – VEE.

To amplify only AC signals, use AC coupling capacitors C1 for the input, and C5 for the output (see Figure 9-24). Alternatively, you can directly couple your output signal via Vout1″. Note the low frequency cut-off is fc 5 1/(2πR1C1) at the input when driven by the low source resistance, Rs, signal such that Rs << R1. For example, many generators have 50Ω source resistances that are << R1 5 20KΩ. For output Vout1, the cut-off frequency is fc 5 1/(2πR4C5).

NOTE: Because the inverting gain op amp circuit has the (+) input tied to a fixed DC voltage within its common mode voltage range, rail to rail output is achieved with any op amp specified with rail to rail output. That is the input range of the op amp does not have to include rail to rail performance. (See Figure 9-24.)

FIGURE 9-24 An inverting gain op amp circuit with AC signal coupling capacitors C1 and C5. Input resistance ~ R1 when C1 is an AC short circuit.

Note that Figures 9-24 and 9-25 look similar with the input resistors R1 and R1″ with R2 and R_BC. The difference is the op amp U1A has very high gain via many transistors inside, and a very low output resistance compared to the amplifier in Figure 9-25.

FIGURE 9-25  A one-transistor amplifier from Chapter 8 that resembles an inverting gain op amp circuit for small voltage gains Av ~ –R_BC/R1″, where R_BC is the feedback resistor. Input resistance ~ R1” when C1 is an AC short circuit.

One can think of Figure 9-25 as simplified inverting gain amplifier. It does have one or two advantages over an op amp. For one advantage, even with typical transis- tors such as the 2N3904, it can deliver very wide bandwidth. For example, if you want to make a cheap inverting gain of –2 amplifier at 1.5 volts supply, you can set RL 5 470Ω, R_BC 5 2KΩ and R1″ 5 1KΩ. The –3 dB bandwidth will be in the order of 5 MHz.

For another advantage, not too many inexpensive op amps work at 11.5 volts supply and have a bandwidth of about 5 MHz for a gain of –2.

In Figure 9-25 the amplifier’s gain is: Vout/Vin ~ –R_BC/R1″ where R_BC/R1″ < 3. However, this simple one-transistor amplifier cannot drive low-resistance loads, and will lose performance in terms of output swing. Also, the maximum gain from this amplifier is less than 3 for distortion less than 1 percent.

But for the flexibility to provide large voltage gain, low distortion, and the ability to load into low-resistance loads with minimal board space area, op amps are the way to go. They are available in even smaller packages such as surface mount ver- sions. These can include SO-8 or SO-14 dual inline surface mount packages, but op amps are also available in even smaller sizes, such as a single amplifier in about the size of an SOT-23 transistor with five leads (e.g., two for power, two for inputs, and 1 for output).

Now let’s take a look at single supply inverting gain op amp circuits. See Figure 9-26.

FIGURE 9-26  An inverting amplifier op amp circuit with a single power supply.

We want to maximize voltage swing by setting the output DC voltage to one-half the supply voltage. This is done by defining a DC voltage at the non-inverting input terminal (e.g., pin 3 for one section of a dual op amp). By negative feedback the volt- ages at the (–) and (1) input terminals pins 2 and 3 are equal, so Vi (1) 5 Vi (–) 5 1VCC/2. Also note that there is a voltage reference filter capacitor C3 that removes noise from 1VCC such as power supply ripple. If there is power supply noise embed- ded into the 1VCC/2 voltage it will be amplified by a factor of [1 1 (R2/R1)]. In this example where R1 5 20KΩ and R2 5 180KΩ, and without C3 any noise from 1VCC/2 will be amplified by [1 1 (180K/20K)] or 10. Thus we want to filter any noise of power line frequency ripple signal that may be present in 1VCC. In general,

a roll-off frequency < 20 Hz is a good start. This is because C3 with R5 and R6 form a low-pass filter.

In Figure 9-26, the low-pass filter roll-off frequency is 1/[2π(R5 || R6) C3] which is: 1/[2π(180KΩ || 180KΩ) 1 μf] or 1/[2π (90KΩ 3 1 μf)] 5 1.77 Hz, which is less than 20 Hz. This means that for any ripple voltage or noise whose frequency is above

1.77 Hz, attenuation will occur. If you need more filtering to remove noise from 1VCC, you can increase C3’s capacitance by using an electrolytic capacitor with its (–) terminal connected to ground. For example, a 33 μf 25-volt capacitor will reduce the noise about 33 times more when compared with the 1 μf for C3.

Since no DC currents flow into R1, and no DC current flows into R2, the DC voltage at output pin 1 is 1VCC/2. See Figures 9-27 and 9-28. Also, if you look care- fully in terms of DC analysis, Figure 9-26 resembles the non-inverting gain amplifier in Figure 9-19.

FIGURE 9-27  Capacitors C1, C3, and C5 are removed for DC analysis.

By removing the capacitors, we see on Figure 9-28 that the circuit is simplified for DC analysis and unconnected components such as R1 and R4 are removed.

Further simplification can be done as shown on the right side drawing of Figure 9-28 where R2 is replaced with a wire such that there is no current flowing through R2, which results in no voltage across R2. And, of course, a wire or 0Ω resistor has no voltage across it. Thus, as we can see on the schematic on Figure 9-28’s right hand side, all we are left with is a voltage follower with 1VCC/2 into the (1) input of the op amp that results in 1VCC/2 at the output pin 1 of U1A.

As a reminder when using a single supply, make sure you have sufficient voltage at the power pin. If you are using general purpose op amps such as LM741, LM4558, TL082, or higher performance ones like the NE5532 or LM4562, usually you can get by with a 1VCC range of 9 volts ≤ 1VCC ≤ 28 volts. However, should you

Tekstvak: 197

FIGURE 9-28 With R1 and R4 removed the circuit is simplified for DC analysis.

decide to use lower-voltage op amps, most will work at 15 volts, but be aware that some of these low-voltage op amps have maximum supply voltages of 15.5 volts (e.g., LMV358) and of 112 volts (e.g., TL972 low-noise op amp has rail to rail out- put voltage).

We can model an inverting gain op amp circuit for AC input and output signals

as shown in Figure 9-29.

FIGURE 9-29  A general model for an inverting gain amplifier.

For an inverting gain amplifier the AC gain G 5 Av 5 –R2/R1 and the AC input resistance Ri 5 R1, with R1 and R2 shown in Figures 9-23, 9-24, and 9-26. In all cases the AC output resistance, Ro ~ 0Ω for the inverting gain op amp circuits.

Troubleshooting Inverting Op Amp Circuits via an Oscilloscope, Signal Generator, and DVM

Here are some tips on basic debugging techniques for inverting gain op amp circuits, which are similar to the non-inverting gain circuits:

1.            Make sure the power supply terminals are connected correctly and use a voltmeter to confirm proper power supply voltages at the power pins. Be sure to have the op amp’s data or spec sheet handy.

2.            Make sure there is at least a ≥ 0.1 μf decoupling capacitor wired or soldered closely to each power pin (e.g., plus supply pin 8 and negative supply pin 4 of an LM1458) and ground. For example, see the 1 μf power supply decoupling capacitors C4 and C3 in Figure 9-23. For a single supply circuit you need only one decoupling capacitor since the other power pin is usually grounded. Without power supply decoupling capacitors, the op amps can oscillate or give noisy output signals on their own.

3.            Measure the voltage with only a DVM across the non-inverting input pin and the inverting input pin of the op amp. Confirm that the voltage is ~ 0 volt. If the voltage is not zero across the two input terminals such as 0.5 volts DC, then the op amp is NOT operating properly.

NOTE: Do not try measuring the voltage across the input pins (e.g., pins 3 and 2 of a dual op amp) with an oscilloscope. You can short out one of the input pins. The ground lead of the oscilloscope probe or wire must always be connected to the ground connection of the circuit. In general, by an indirect connection, the ground lead of an oscilloscope is at ground potential. That is, the ground lead from an oscilloscope is not “floating” like the negative (and positive) lead of a DVM.

4.            For op amp circuits with positive and negative supplies, verify that the DC output at the op amp voltage is close to zero. If it is not, verify that the non- inverting input terminal is connected to a DC potential such as ground.

5.            If after #1 to #4, you are not getting anywhere, replace the op amp, but make sure the original op amp could operate at the power supply voltages. For example, if the power supply provides ± 12 volts DC and the circuit does not seem to work, verify that op amp is not a lower-voltage version, such as an LMV358 (e.g., 15.5 volts maximum) or a TLC272, where a supply voltage of more than ± 8 volts will cause damage.

6.            If the DC bias conditions are correct, but the AC signals are distorted or do not have correct amplitudes given the gain setting [–R2/R1], then first check the resistance values of the feedback resistors. Also make sure the feedback network resistance value of R2 ≥ 500Ω is not low enough to cause current limiting at the output that will cause clipped waveforms.

7.            If the low-frequency (e.g., 20 Hz) response is poor, such as less than 50 percent of amplitude when compared to the signal amplitude at a mid-band frequency (e.g., 1000 Hz), check the input and output coupling capacitors. If they are OK, check the frequency response at the output pin of the op amp such as pin 1 or 7 of a dual op amp. If there is a loss in low-frequency response, see if there is an AC coupling capacitor at the input, such as C1 in Figure 9-24 or C1 in Figure 9-26. If you suspect there is a problem like the capacitance is not high enough, parallel a larger capacitance across C1, such as a 2.2-μf capacitor. If the capacitor is ceramic, check the maximum working voltage. For example, if you use a 1-μf 6.3-volt capacitor and the DC voltage across it is 15 volts, chances are that the true capacitance will be half of that, or 0.5 μf, which will cause a poorer low-frequency response.

8.            Also, if electrolytic capacitors are used, measure with a voltage meter that the voltage across the capacitor is correct with the polarity markings of the capacitor. If you measure a reverse bias voltage across an electrolytic capacitor, turn off the power immediately, replace the capacitor, and install a new capacitor with the correct polarity.

9.            Check to make sure the phase at the output is inverted relative to the phase of the input signal. If you are getting the same phase signal at input and output, then the op amp may be damaged or open circuited from the feedback and input resistors (e.g., R2 and R1). As a result of the op amp being open circuited,

the signal will travel through R1, then to R2 to the output. If you have no power to the op amp, a feed-through condition will also occur and the signal from R1 to R2 may travel through and there will be no inversion of the input signal at the output terminal.

A telltale sign that something is wrong is if you probe the inverting input of the op amp (e.g., pin 2 or pin 6 of a dual op amp) and see a signal, then something is wrong because you should see no signal at all. This is because the inverting input in an inverting gain amplifier is a virtual ground, or a vir- tual AC ground (e.g., in a single supply circuit).

10.            Also check if the circuit was for a dual op amp but a single op amp was installed instead, and vice versa. Both single and dual op amps have 8 pins total and some pins are common, such as the –V supply pin 4, non-inverting input pin 3, and inverting input pin 2; however, the other pins 1, 5, 6, 7, and 8 are not the same for a single op amp and dual op amp.

11.            In an inverting gain amplifier you cannot make the feedback resistor R2 5 0Ω, since this will result in a zero output. Also, making the feedback resistor (e.g., R2) low resistance (e.g., < 500Ω) can cause the op amp’s output waveform to distort due to current limiting. Recall that the feedback resistor is also connected to the inverting input terminal, which is a virtual ground. So, the feedback resistor is a “direct” load resistor to the output terminal.

Brief Notes on Both Non-inverting and Inverting Gain Amplifier Circuits

•     If you are providing signals to the outside world, such as to a connector that will later be used with a cable or wire, please add a 47Ω series resistor between the output of the op amp and the connector. Or if you are driving any shielded cable or coaxial cable with the op amp, also add a series resistor between the output of the op amp and the cable. The reason for adding a series resistor is to prevent the op amp from oscillating, and to provide some short circuit protection. This is especially true when a person can inadvertently short out a connector while trying to connect a cable to it.

•     For both types of amplifier circuits, choose an op amp that is “unity gain compensated.” This will ensure that there will be no parasitic oscillations. Most op amps are unity gain compensated such as the ones listed in this chapter.

•     The high-frequency bandwidth is determined by the op amp’s unity gain bandwidth divided by [1 1 (R2/R1)]. Usually, there is a typical or minimum unity gain bandwidth specification on the data sheet. For example, if you are using a TLO82 op amp that has about a 3-MHz unity gain bandwidth, the high-frequency bandwidth of either inverting or non-inverting amplifiers is determined by the ratio of R2 over R1. In a previous example R1 5 2KΩ and R2 5 18KΩ. Therefore, with a TL082 the high-frequency bandwidth is

3 MHz/[1 1 (18KΩ/2KΩ)] 5 3 MHz/10 or 300 kHz. This 300-kHz

bandwidth is for a non-inverting gain of 110 or an inverting gain of –9.

•     The op amps also have a power bandwidth specification or slew rate specification. Generally, for instrumentation such as monitoring slow changing DC voltages, a slew rate of ≥ 0.2 volts per microsecond will work. For audio circuits with ≤ 5 volts peak to peak output, ≥ 0.5 volts per microsecond is fine. But for a large audio signal voltage output of ~ 24 volts peak to peak, ≥ 1 volt per microsecond will do. Once you have to work with signals beyond 20 kHz, it’s best to find an op amp ≥ 20 volts per microsecond. For example, for even standard resolution TV (SDTV) analog signals such as 480i, you will need ≥ 50 volts per microsecond with a high-frequency bandwidth ≥ 5 MHz.

•     Op amps with negative feedback, such the non-inverting and inverting amplifiers, have power supply noise rejection at their power pins. This means if there is a power supply ripple from a raw supply, the op amp will keep the ripple from showing up at the output. However, it is usually better to use a voltage regulator for powering op amps. See the next section in this chapter.

A Short Look at Linear Voltage Regulators

Voltage regulators provide a constant voltage to an electronic circuit, even though the raw power supply is varying in DC voltages. For linear voltage regulators to oper- ate correctly, the incoming voltage must be larger than the regulated voltage at the output. For example, if you have a power supply that varies between 9 volts and 12 volts due to power supply ripple or changes in the AC power line voltage, you can provide a regulated 16 volts or 15 volts via an LM7806 or LM7805 three-terminal regulator. If you want to adjust to different regulated voltage, you can use an LM317T integrated circuit and two resistors to set the voltage, which will be greater than two volts difference between the incoming voltage and the regulated output voltage.

Another reason for using linear voltage regulators is to remove high-frequency noise from switching power supplies. For example, an efficient wall power supply may have switching noise in the 1-MHz region added to its regulated DC voltage. A voltage regulator can remove the switching noise.

Voltage regulators are available in through-hole and surface mount packages. We will be looking at through-hole versions, which are easier to use for DIY projects, plus they can be mounted onto heat sinks.

Types of Linear Regulators Starting with “Standard” Fixed Voltage Versions

What is a voltage regulator? It is a device that takes in an incoming raw DC voltage and provides a constant lower voltage. The incoming raw DC voltage may vary in DC amplitude with noise or ripple voltages riding on top of a DC voltage. The volt-

age regulator then removes the noise or ripple voltages and provides a lower constant voltage.

For example, if the incoming DC voltage has a range of 17 volts to 110 volts, that is, a minimum of 17 volts and a maximum of 110 volts, and this incoming DC voltage is connected to a 15 volt regulator, the output of the regulator will be a con- stant 15 volts.

A voltage regulator requires a DC input voltage that does not dip below its min- imum input voltage rating. This means that the DC input voltage must not be a pulsating DC voltage such as a rectified AC voltage without a filter capacitor, which means the DC voltage can dip to 0 volts. If such an instance happens, the regulator will just pass through the input voltage that includes the pulsations. Figure 9-30 shows an example of a pulsating DC voltage waveform.

FIGURE 9-30 A half wave rectified voltage without a filter capacitor that is NOT the type of DC input voltage required for a voltage regulator.

The waveform at times goes to 0 volts or ground. Instead, see Figure 9-31, a half wave rectified voltage with a filter capacitor that is more suitable for use with a volt- age regulator.

FIGURE 9-31 Top trace: A half wave rectified voltage that provides a DC voltage with a “saw-tooth” ripple voltage riding on top of the average DC voltage for the input voltage into a voltage regulator. Bottom trace: Regulated voltage output. The center horizontal line is at 0 volts.

Note the wave does not go to ground and has a minimum DC voltage at the low- est point of the sawtooth. This lowest point voltage must be equal to or greater than the “drop-out” voltage specification of the voltage regulator. For example, if Figure 9-31 is at 2 volts per division, the DC voltage has a minimum voltage of about 14 volts and a maximum voltage of about 2.8 div 3 2 volts/div, or about 5.6 volts peak. If a standard voltage 3.3-volt voltage regulator is used, generally, the drop-out volt- age is about 2 volts to be on the safe side, which would mean a minimum input volt- age 5 3.3 volts 1 Vdrop-out is needed. This would be 3.3 v 1 2 v 5 5.3 volts, which would mean when the waveform dips below 5.3 volts, the voltage regulator will stop working properly since the example has the waveform at a peak of 5.6 volts only briefly and dipping to down 4.0 volts. To “fix” this, you can use a low–drop-out regulator that usually has a drop-out voltage of about 0.5 volts. This means the min- imum input voltage 5 3.3 volts 1 0.5 volt or 3.8 volts. Since the example waveform has the minimum voltage at 4.0 volts, the low–drop-out voltage regulator will work properly, but just barely.

Drop-Out Voltage Summary

Again, most standard linear voltage regulators (e.g., with fixed and adjustable regu- lated output voltages) require that the “dip” or minimum of the DC input voltage is at least 2 volts above the regulated output voltage. The drop-out voltage is specified as the minimum voltage between the input and output of the regulator to ensure that

voltage regulation takes place. In the example of having 2 volts above the regulated voltage, the drop-out voltage is 2 volts.

However, there are low–drop-out voltage linear voltage regulators that can regu- late the voltage when the “dip” or minimum of the DC input voltage is at least 0.50 volt above the regulated output voltage. In this case, the drop-out voltage is 0.5 volts, which when compared to the 2-volt drop output voltage for standard voltage regula- tors, we can see why the term “low–drop-out voltage” is used.

Voltage Selections, Packages, Pin Outs and Schematics

For standard fixed-voltage output regulators in TO-220 (1 amp) and TO-92 (100 mA) packages, see Figure 9-32. Notice that the pin outs are not the same for positive and negative voltage regulators.

FIGURE 9-32 Fixed-output voltage regulators shown with 1 amp TO-220 and their mounting tabs, and the smaller 100 mA TO-92 packages that look like transistors (e.g., 2N3904), but they are actually integrated circuits.

This is important to notice that in a positive regulator, reading from the side where the part number is printed, the sequence is input, ground, and output, whereas in a negative voltage regulator, the first two terminals are reversed have the sequence, ground, input, and output. For TO-220 packages, the metal tab is always connected to the middle terminal. That is in a TO-220 fixed-voltage regulator, the metal (mount- ing) tab is ground for a positive version, while for a negative fixed-voltage regulator the tab is the input voltage terminal. It is very important to never ground the negative voltage regulator’s tab, or you will short out the raw input DC supply.

If you are using a low-power TO-92 linear voltage regulator, such as the positive voltage 78Lxx or negative voltage type 79Lxx, be aware of the pin out because they are different. Again, from the flat side where you are reading the part number, the sequence for the positive TO-92 package is output, ground, and input, which is a reverse sequence from its bigger TO-220 version. For the negative voltage regulator in TO-92 package, the pin out sequence as read from the flat side is ground, input, and output, which is the same as its TO-220 higher current version. Keeping track of all these packages in your head can be confusing. If in doubt, look up the data sheet.

See a schematic diagram for fixed-output voltage regulators in Figure 9-33 and their associated decoupling capacitors.

FIGURE 9-33 Schematic diagrams for standard positive and negative voltage TO-220 regulators with decoupling capacitors C1, C2, C3, and C4. Notice that the first two pins U1 are reversed from the pins in U2 (Vi and Gnd).

Knowing the Pin Out Sequence Is Important

See Figure 9-34 for the pin out sequence for both TO-220 and TO-92 fixed (e.g., not adjustable) voltage regulators.

Common fixed output voltages include 13.3 volts (78Mxx for 500 mA or 78Lxx for 100 mA), with 78xx (positive) for 1 amp at 15 volts, 16 volts, 18 volts, 110 volts, 112 volts, 115 volts, 118 volts, and 124 volts. In the 78xx part number, you replace the “xx” with the voltage you require. For example, if you need a 112-volt regulator 78xx  7812, or if you need a 15-volt regulator, the correct part number is 7805.

Equivalently, there are LM340 series regulators equivalent to the 78xx series. Here, the numbering system requires a dash such as a 15-volt regulator is named as LM340-5. Voltage selection on the LM340 series is the same for the 78xx series when using parts made by Texas Instruments (TI).

FIGURE 9-34 Pin out sequences as read from the front side of the voltage regulators that have their printed markings. The TO-92 illustrations are drawn larger for clarity of pin outs. See Figure 9-32 for a comparison of the real sizes of the different packages. (7800 series = Positive regulator, 7900 series = Negative regulator).

For negative voltage regulators, the 79xx series can output regulated negative voltages of –5 volts, –6 volts, –12 volts, –15 volts, –18 volts, and –24 volts. Again, as an example, if you need a –6-volt regulator, the part number will be 7906.

NOTE: The raw input DC volts should not exceed any voltage higher than 30 volts at its peak for positive voltage regulators (78xx) and no lower than – 30 volts peak for negative voltage regulators (79xx); otherwise, they can be damaged.

Low–Drop-Out Voltage Regulators

There are low–drop-out (LDO) voltage regulators that require that the input voltage is only ≥ 0.5 volt above the regulated output voltage. Again, this means the drop-out voltage is 0.5 volt. Also, these LDO regulators have the same pin out sequence as the previously mentioned standard fixed regulators. That is, the pin outs for the low– drop-out positive voltage regulator LM2941-x are the same as the LM78xx TO-220 regulators. The low–drop-out negative voltage regulator, LM2990-xx, will have the same pin out for the LM79xx TO-220 packages. See Figure 9-35.

FIGURE 9-35 Low–drop-out voltage regulators and their pin outs. (LM2940 series = Positive regulator, LM2990 series = Negative regulator).

However, there are a couple of noteworthy differences between the low–drop- out regulators compared to the standard ones.

•     The maximum positive input voltage is 125 volts for the LDO regulators versus 135 volts absolute maximum for the standard versions (LM78xx).

•     The decoupling capacitors have to be larger for the output capacitors in low– drop-out regulators. Typically, you can use electrolytic capacitors of 33 μf at 35 volts for both the input and output decoupling capacitors, and most of all observe the polarity. For the positive LDO regulators, a 1 μf (e.g., ceramic) at the input terminal will suffice while using a 33 μf at the output terminal. And with the LM2990-x series, the negative terminals of the electrolytic decoupling capacitors are connected to the input and output terminals of the regulator. See Figure 9-36 and observe the electrolytic capacitors’ (C2, C3, and C4) polarity markings.

The available fixed-output voltages for the positive voltage LDO regulators LM2940T-xx are: 5 volts, 8 volts, 9 volts, 10 volts, 12 volts, and 15 volts. For example, a positive 5-volt version would be listed as LM2941T-5.0, whereas a 9-volt version would be LM2940T-9.0, and a 12-volt part number will be LM2940T-12.

For the negative voltage LDO regulator LM2990T-xx, the available fixed-output voltages are: –5 volts, –12 volts, and –15 volts. An example part number for a –5-volt version would be listed as LM2990T-5.0, whereas a negative 12-volt version would be LM2990T-12.

FIGURE 9-36 LDO regulators with decoupling capacitors larger in capacitance for C2, C3, and C4 when compared to standard voltage regulators.

Things That Can Go Wrong with 7800 and 7900 Series Voltage Regulators

•     If the input voltage to a positive voltage regulator (e.g., 7805) is a negative voltage that is the wrong polarity, the positive voltage regulator will drain excessive current and the output will be about –0.9 volts. You will need to shut off power immediately and replace the regulator with a new positive voltage regulator. You then connect with the correct polarity input voltage (e.g., a positive voltage) to the input of the new positive voltage regulator.

•     If the input voltage is correctly positive, but the ground lead (e.g., middle lead) is left unconnected (e.g., floating), the output voltage will be roughly the same voltage as the input. This can lead to damaging ICs or circuits that require a regulated voltage. For example, suppose you have a TLC272 op

amp that is connected to a 112-volt regulator’s output, 7812, but the 7812’s middle ground lead is not connected to ground. With an input voltage of 124 volts, the output voltage of the 7812 will be about 23 volts, which will damage the TLC272 op amp because it has a maximum operating voltage of 16 volts.

•     If you reverse the pin outs by swapping the input and output leads of a positive voltage regulator (e.g., 7805) while still grounding the middle lead, then you will be applying the unregulated voltage to the output pin and taking the output of the regulator via its input pin. This will cause excessive current and also pass the unregulated voltage from the regulator’s output pin to the regulator’s input pin. In other words, you will be applying an unregulated high voltage to your circuits, which can damage them. Swapping the positive voltage regulators’ two outer leads can be the result of not paying attention to the silkscreen on a printed circuit board that denotes which way the TO-220 package is installed. Power should be turned off immediately and the regulator replaced and installed properly.

•     If the input voltage to a negative voltage regulator (e.g., 7905) is an incorrect positive voltage of the wrong polarity, the negative voltage regulator will drain excessive current and the output will be about 12 volts. You will need to shut off power immediately and replace the regulator. Then fix the wiring of the input voltage source such that it provides a negative voltage to the input pin of the negative voltage regulator.

•     If the input voltage is correctly a negative voltage, but the ground lead (e.g., first lead on the left side) is unconnected (e.g., floating), the output voltage will be roughly the same negative voltage as the input’s negative voltage. Hence the raw input voltage is passed onto the circuits requiring a well- defined or limited negative voltage. This can lead to damaging ICs or circuits that require a regulated negative voltage.

•     If you reverse the pin outs by swapping the pin sequence from 1, 2, 3 to 3, 2, 1 of a negative voltage regulator (e.g., 7905), the output lead will be grounded, the input lead stays the same since it is the middle pin, and output is now provided via the ground pin. This will cause excessive current and output of about –0.7 volt. You will need to shut off power immediately and replace the regulator with the correct pin out sequence.

Accidents Involving the 7800 Series Positive and 7900 Series Negative Voltage Regulators

If you load onto a printed circuit board a positive 78xx voltage regulator in the place of a negative 79xx voltage regulator and 79xx regulator in the place of a 78xx voltage regulator, you will have fed through the raw voltage with a voltage drop of about the rated regulated voltage.

210    C h a p t er Ni n e

For example, if 7808 is installed in the place of a negative voltage regulator, 79xx, and the raw supply is minus 20 volts, then the output voltage will be –(raw voltage

– rated regulated voltage), which is then: –(20 volts – 8 volts) 5 –12 volts.

And if a negative voltage regulator such as a 7906 is installed where a positive voltage 78xx regulator should be, the output voltage will be (raw voltage – rated regulated voltage) 5 (20 volts – 6 volts) 5 114 volts.

If the raw positive voltage is equal to or less than the regulated voltage rating, the output will be close to zero for the 7900 negative series regulator that is loaded into a board that is meant for a 7800 series regulator.

Some Mishaps with Low–Drop-Out LM2940 and LM2990 Series Regulators

The low–drop-out (LDO) regulators have different internal circuitry as the standard regulator circuits. Thus, there will be some differences when LDO regulators are connected incorrectly.

•     When a positive voltage LDO regulator (e.g., LM2940-xx) is connected with the input and output leads swapped, and a correct raw positive voltage is applied, the regulator will pass through the raw positive voltage. For example, if the raw voltage is 120 volts DC, and the input and output leads of the LDO positive voltage regulator is swapped, the output voltage will be about 120 volts. Therefore, there will be no regulation, and subsequent circuits requiring a regulated voltage may be damaged. You should turn off the raw power supply and replace the LDO and install a new one with the correct pin out sequence.

•     If you apply a negative input voltage to a positive voltage LDO’s input terminal, the output will be close to zero volts.

•     If you forget to connect the middle terminal of the positive voltage LDO to ground and leave the middle pin unconnected or floating, again the output will be essentially the unregulated input voltage that can cause damage to other circuits.

•     If you have inadvertently installed a negative LDO (e.g., LM2990-xx) in place of a positive voltage regulator, and turned on the positive raw supply, the negative voltage LDO will pass to its output the raw input supply voltage. This raw input voltage will be higher than the expected regulated voltage and can damage other circuits.

•     For a negative LDO voltage regulator (e.g., LM2990-xx), if you apply an incorrect positive voltage at the input, the output will be about 11 volt from the negative LDO regulator, and excessive current will draw. The raw power supply must be turned off immediately. Then the correct negative supply must be wired to the input terminal.

•     If you reverse the pin sequence of the negative LDO voltage regulator, excessive current will be drained and about –0.7 volt will be at the output. Again, turn off the supply immediately and install a new negative LDO voltage regulator with the correct pin out sequence.

•     If the circuit requires a negative LDO voltage regulator and everything is connected correctly, but a positive LDO voltage regulator is mistakenly installed, the raw supply voltage will pass to the output and possibly cause overvoltage damage to other circuits.

Some Troubleshooting Hints When Working with Regulators

•     One thing to remember about voltage regulator chips is that they can only deliver current in one direction. In a positive regulator, the current flows out. But if you try to have the positive voltage regulator absorb or sink current, you may turn off the voltage regulator and as a result, the output voltage will no longer be the expected regulated voltage. An example is if you have a 15-volt regulator (7805) and if one end of a resistor is connected to the output of the 7805 and the other end of the resistor is connected to 120-volt source, then current will be flowing the wrong way into the output terminal. This can cause the 7805 to shut off because it can only absorb a small amount of current.

•     If the voltage regulators get too hot, then they will shut down and output close to zero volts. You will need to first check if you have the correct voltage regulator installed accordingly. Then you need to monitor the current into the voltage regulator’s input terminal via an ammeter. See if it’s within the range of the output current capability of the voltage regulator. If it is, then the voltage regulator may need to be mounted on a heat sink.

•     If the output is lower than the regulated voltage, use a DVM and measure the voltage at the input and output terminals. For a positive voltage regulator, the input voltage should be greater than 2 volts above the rated output voltage. For example, if you are using a 7808 for 18 volts output, the minimum input voltage is 2 volts above that, which means that the input voltage is greater than or equal to 10 volts.

•     If you are monitoring with an oscilloscope on the input and output terminals and there is a high-frequency oscillation signal, then make sure you have the decoupling capacitors soldered or wired within one-half inch of the input, ground, and output terminals. Also make sure the capacitances are large enough (≥ 1 μf ).

If you are using ceramic capacitors, use a voltage rating at least twice the voltage at the input and output terminals. This is because some ceramic capacitors will de- rate when there is a DC voltage close to its maximum rating.

Should an electrolytic capacitor be used, observe its polarity. If it is installed incorrectly, shut off the power immediately and install a new one with the correct polarity.

This completes Chapter 9, and we will be looking into audio circuits in the next chapter.

            CHAPTER 10

Audio Circuits

We will explore two types of audio amplifiers—preamplifiers, commonly known as preamps, and power amplifiers.

The examples chosen will be from commercially manufactured designs. What’s different in this chapter is that in some cases, we will see associated circuits such as a power supply and a voltage regulator. Although integrated circuit (IC) amplifiers could have been chosen for this chapter, it would be good to know something more about amplifiers on the transistor level. This way you can use this knowledge to aug- ment IC designs with discrete circuits.

NOTE: Audio signals are usually measured in RMS voltages. For a sine wave signal, the peak to peak voltage is 2.82 x RMS voltage. For example, a 1-volt RMS sine wave is 2.82 volts peak to peak. However, when you are measuring the audio signal on an oscilloscope, you will often measure the peak to peak amplitude instead of the RMS value.

Preamps and Power Amps

For transducers such as microphones and phonograph pickup cartridges that gener- ate small amplitude signals in the 1 mV to 10 mV range, preamps are needed to provide large voltage gains for these devices. For example, a microphone preamp has a voltage gain of 30 to 1000.

A dynamic microphone needs a preamp gain of about 300 or 50 dB. There are higher output microphones such as electret types. These require power, and they gen- erally have about 10 times more output signal than a dynamic microphone; so they only require gains of 30 or about 30 dB. Low-level microphone preamps have voltage gains from about 30 to 1000 (30 dB to 60 dB) with flat frequency response from 20 Hz to 20,000 Hz.


NOTE: Decibels or dB = 20 log10(|Vout/Vin|) where the voltage gain = |Vout/Vin|. For example if you have a 2 volt RMS input signal and the output is 10 volts RMS, then the gain = |10 volts RMS/2 volts RMS| = 5 and in decibels we have 20log10(5) dB = 14 dB since log10(5) = 0.7. Also log10(x) is commonly known as just log(x), where is a non-negative number. For example, ≥ 0.

Magnetic phonograph pick up cartridges (a.k.a., phono cartridges such as Shure M97xE or Audio Technica AT95E) for playing (LP) vinyl records generally require a gain of 50 to 100 at 1 kHz. The phono cartridge preamp requires a frequency equal- ization to provide flat frequency response from the vinyl record to the output of the preamp. That is, the vinyl recording is cut with a non-flat frequency response. This is done to maximize playing time by essentially attenuating low frequencies while boosting higher frequencies.

For higher (amplitude) level signals, we use line-level preamps that have a gain of 1 to about 10 or 0 dB to 20 dB. Line-level signals are in the order of 70 mV RMS to about 1 volt RMS. A typical portable digital audio device may output a maximum of about 1 volt RMS or about 2.82 volts peak to peak. Preamplifiers, whether low- level or line-level types, do not provide enough current output to drive a loudspeaker. However, some preamps can drive headphones.

The ultimate goal for the preamp is to provide enough signal level to a power amplifier. Typically, an audio power amplifier will be specified for an input voltage for full power output.

For example, many audio power amps have a gain of 26 dB or 20. If the amplifier is specified for 2 volts RMS input for full output into an 8Ω speaker load, then the power output will be output RMS voltage squared, divided by the load resistance. In this example, power output will then be P 5 [(2 volts RMS 3 20)2/8 Ω]. Thus, P 5 [402/8] watts 5 [1600/8] watts, or P 5 200 watts output into the 8Ω load.

For this chapter we will look into a Dynaco PAT-5 low-level preamp and two dif- ferent power amplifiers from the Pioneer SX-626 and SX-636 stereo receivers.

First, we need to discuss differencing amplifiers and emitter follower amplifiers and how they will relate to preamps, power amps, or voltage regulators.

Now let’s look at some building blocks to preamp and power amps.

A Basic Difference Amplifier

In most modern amplifiers, negative feedback is used. This requires an inverting input terminal, and a non-inverting input terminal. For example, the voltage fol- lower circuit using an op amp has the output terminal connected to the inverting input terminal while the input signal is applied to the non-inverting input terminal (e.g., Chapter 9, Figure 9-3).

A difference amplifier can be made with a common emitter amplifier and a com- mon base amplifier. What we want is for the difference amplifier to provide an out- put signal that somehow subtracts two input signals. First let’s take a look at each amplifier in Figure 10-1.

FIGURE 10-1 A common emitter amplifier with input signal Vin1 and a common base amplifier with a second input signal Vin2.

The common emitter amplifier in Figure 10-1 provides an inverting gain refer- enced to input signal Vin1. In both figures, VE 5 emitter voltage, VB 5 base voltage, IC 5 collector current, Rin1 5 input AC resistance Q1’s base, and Rin2 5 input AC resistance into Q1’s emitter.

Generally for the common emitter amplifier, the gain – k 5 –R3/re, where re 5 0.026 v/IC, where in this case, IC 5 (VE – –12 v)/R2, and where VE ~ –0.7 volt since Q1’s DC base voltage ~ 0 volt.

Therefore, IC 5 (–0.7 v – –12 v)/20KΩ or IC 5 11.3 v/20KΩ 5 0.565 mA or

IC 5 0.565 mA 5 0.000565 A, and re 5 0.026 v/0.000565 A or re 5 46Ω.

With re 5 46Ω, the gain, Vout/Vin1 5 –k is then –R3/re or –k 5 –10KΩ/46Ω

Vout/Vin1 –k –217.

If –k 5 –217, then by multiplying by –1 on both sides, 217. Knowing 1k will be important for the common base amplifier shown in Figure 10-1.

The input resistance looking into Q1’s base, Rin1 is β re, and if β 5 100 (as an example only because β can typically be anywhere from 50 to 300), then Rin1 5 100 (46Ω) or Rin1 5 4600Ω.

The actual input resistance as seen by Vin1 is then R1 || Rin1, and in this case, R1 || Rin1 5 20KΩ || 4600Ω 5 3740Ω.

NOTE: R1 || Rin1 = (R1 x Rin1)/(R1 + Rin1).

Now let’s look again at Figure 10-1’s common base amplifier on the right side. The base of the Q2 is AC grounded via C1. And we call it a common base amplifier because the base is AC grounded and both input and output signals have ground connections. Here we see that the common base amplifier has no phase inversion at the output with the collector output voltage being Vout 5 1k (Vin2).

It turns out a common base amplifier, when the input signal Vin2 is a voltage source with 0Ω source resistance, has the same voltage gain as the common emitter amplifier except there is no phase inversion at the output. That is, the common base amplifier’s output waveform has the same phase as the input signal. Given the same collector load and emitter bias resistors R3 and R2 with the same power supplies at

± 12 volts, the collector current of the common base amplifier for Q2 is also 11.3 volts/ R3 or 11.3 volts/20KΩ 5 0.565 mA 5 IC. Thus re 5 0.026 volt/0.000565 A, or re 5 46Ω is the same for the common emitter amplifier in Figure 10-1.

The gain of the common base amplifier Vout/Vin2 5 1 k 5 R2/re 5 10KΩ/46Ω

5 1217, which is the same gain of the common emitter amplifier except that there

is no phase inversion in a common base amplifier between the input and output signals.

When we look into the input resistance via Rin2 for a common base amplifier, we find that there is no current gain in terms of input emitter current and output collector current since they are essentially the same. However, in the common emit- ter amplifier, the base current is very small and is 1/β of the collector current. Thus, we can surmise that since there is no current gain in the common base amplifier, the input resistance must be very low. In fact, the input resistance into the emitter of Q2 is Rin2 5 re 5 46Ω, which smaller than Rin1 (input resistance into Q1’s base) by a factor of 1/β since Rin1 5 β re. Recall that if β 5 100, then Rin1 5 4600Ω for the common emitter amplifier. But Rin2 for the common base amplifier is 46Ω. The actual input resistance from Vin2’s viewpoint is R2 || Rin2 5 20KΩ || 46Ω ~ 46Ω.

NOTE: Anytime you parallel a first resistor with second resistor of at least 35 times more resistance, the paralleled resistance is essentially the value of the first resistor within 3 percent accuracy. For example, a 1000Ω resistor in parallel with a 39KΩ is about 975Ω, which for most practical purposes is still 1000Ω.

In Figure 10-1, we now have shown that a one-transistor circuit can have two inputs, one at the base and another at the emitter. By utilizing two inputs with a single transistor, we can make a difference amplifier as shown in Figure 10-2.

Since the amplification factor, k 5 R3/re, is the same for both common emitter and common base configurations, we see in Figure 10-2, a difference amplifier is possible with a single transistor Q1. The output signal Vout 5 k (Vin2 – Vin1).

Or put another way, for the common emitter amplifier’s input signal Vin1, Vout 5 –k (Vin1); and for the common base amplifier, Vout 5 k (Vin2).

FIGURE 10-2 A difference amplifier having two input signals, Vin1 and Vin2, and a block diagram of the difference amplifier on the right side.

We can use this difference amplifier with another common emitter amplifier to make a feedback amplifier as shown in Figure 10-3.

FIGURE 10-3 An example two-stage feedback amplifier with the first stage transistor with Q1 similar to the difference amplifier shown in Figure 10-2.

Let’s first find the DC bias voltages. VB1 5 10 volts DC since R3 and R2 form a divide by two voltage divider circuit for 120 volts. For an estimate as long as R2 || R3 ≤ 10 3 R4, we can ignore base currents as long as β >> 10 (e.g., the β > 300 for both Q1 and Q2).

With VBEQ1 5 VB1 – VE1, VE1 5 VB1 – VBEQ1 or VE1 5 10 v –0.7 v or VE1 5

9.3 volts. From this, Q1’s DC collector current ICQ1 5 IEQ1 5 9.3 v/20KΩ 5 0.465 mA. Thus, VC1 5 20 v – ICQ1 3 R5 5 20 v – 0.465 mA 3 10KΩ or VC1 5 20 v – 4.65 v, which leads to VC1 5 15.35 volts.

Note that although we are not optimized for maximum swing, it will be fine because the voltage gain of the next stage is very high and less than 1 volt peak to peak swing from Q1’s collector will do. Also note VCEQ1 5 VC1 – VE1 5 15.35 v –

9.3 v 5 16.05 volts, which is good since the collector to emitter voltage should typi-

cally > 0.5 volt to ensure that Q1 is operating as an amplifying device. For Q2, the base voltage is VB2 5 20 v [R8/(R8 1 R7)], and thus:

VB2 5 20 v [20K/(20K 1 180K)] 5 20 v [20K/200K] 5 20 v /10

VB2 5 2 volts, and since VBEQ2 5 0.7 volt

VE2 5 VB2 – VBEQ2 5 VE2 – 0.7 volt 5 1.3 volts or VE2 5 1.3 volts

This means ICQ2 5 IEQ2 5 VE2/R10 5 1.3 v/1300Ω or ICQ2 5 1 mA.

VC2 5 20 v – ICQ2 R11 5 20 v – 1 mA (10KΩ) 5 20 v – 10 v

VC2 5 110 volts

A single-supply two-transistor amplifier utilizes a difference amplifier Q1 where the emitter acts as a second input terminal for the feedback resistor network, R9 and (R6 || R4). Q1’s base can be thought of as a first input terminal. The difference volt- age across the base-emitter junction of Q1 provides an output signal at the collector, VC1. This output signal is then sent to the second-stage amplifier, Q2, which is another inverting gain amplifier.

To understand this amplifier in terms of a feedback system, we can find which terminal of Q1 is the non-inverting input and which one is the inverting input. One way is to look at Q1 and Q2. We know that Q1 is a common emitter amplifier with phase inversion from the base to the collector in terms of AC signals (e.g., see Figures 10-1 left side and 10-3). Since the second stage with Q2 is also a common emitter amplifier, it inverts the phase from Q1’s collector signal. Thus, from the base of Q1 to the collector of Q2, we get two 180-degree phase inversions so that the output signal at Vout is in phase with the input signal Vin.

For Figure 10-3’s amplifier, we can say that Q1’s base terminal is a non-inverting input. By process of elimination, is the remaining input at the emitter the inverting input for this two-transistor amplifier? Yes, that is the case. If we couple an input emitter terminal AC signal at the emitter of Q1 via C3, the collector voltage VC1 will have an AC signal in the same phase with this emitter terminal input signal. Since Q1’s collector signal is sent to Q2, which is a common emitter amplifier, there will be a phase inversion from the collector signal of Q1 to the collector signal of Q2.

In other words, since a signal coupled to the emitter terminal at Q1 has the same phase as the collector signal at Q1, the phase is thus inverted from the Q1 emitter terminal signal to the collector of Q2, which is coupled to Vout. Thus, the emitter terminal of Q1 is an inverting input terminal for this amplifier.

We can estimate the non-inverting gain by just observing that the two-resistor feedback network R9 and R6 looks like the feedback network used in non-inverting gain op amp circuits. For a more accurate estimate we can replace R6’s value with R6 || R4.

However, since R4 (20KΩ) >> R6 (510Ω) we can say that (R4 || R6) ~ R6 5 510Ω.

The estimated (closed loop) gain is then [1 1 (R9/R6)] ~ Vout/Vin, providing the total (open loop) gain from the first and second stages is >> [1 1 (R9/R6)].

For this example, R9 5 100KΩ and R6 5 510Ω, so our estimated gain is Vout/Vin ~ [1 1 (100K/510)] or Vout / Vin ~ 197.

The quick analysis of the gain of the first stage involving Q1 is about:

(R5 || R7 || R8 || Rin2) / R6, Given R5 5 10KΩ, R7 5 180KΩ, R8 5 20KΩ, and

Rin2 ~ 10KΩ when Q2’s β ~ 400 at ICQ2 5 1 mA collector current

To calculate Rin2, with ICQ2 5 1 mA, re2 5 0.026 v/ICQ2 5 0.026 v/0.001 A 5

26Ω. Or re2 5 26Ω. Rin2 5 β re2 5 400 3 26Ω or Rin2 5 10.4kΩ or Rin2 ~ 10kΩ. The gain of the first stage is about (R5 || R7 || R8 || Rin2)/R6 ~ 4K/510 ~ 8.

Note VB2 ~ 2 volts, so VE2 ~ 1.3 volts and with R10 5 1300Ω, the emitter and collector currents of Q2 is ~ 1.3 volts/1300Ω or ~ 1 mA 5 IC2. The second stage’s gain from the base of Q2 to its collector is about R11/re2, where R11 10kΩ with re2 5 0.026 volt/IC2 re2 5 (0.026 volt) / (0.001 A) or re2 26Ω.

Gain of the second stage 5 ~ 10KΩ/26Ω or about 380.

Total open loop gain is then 5 gain of 1st stage gain of 2nd stage 5 8 3 380

~3040. The estimated “closed loop” gain is ~ 197 and so we should be OK since 3040

>> 197. Should the open loop gain be in the order of [1 1 (100K/510)], such as the gain of 1st stage 3 gain of 2nd stage 5 197, the actual gain Vout/Vin will be less than than [1 1 (100K/510)] 5 197.

Also note that the gain calculations are estimates, by which we mean if the mea- sured gain is within 20 percent of the calculated value, that’s good enough. Estimates within 20 percent are good enough for troubleshooting. If more accurate analysis is needed probably the best thing to do is to run a simulation via a program such as LT Spice if you are not into deep engineering analysis via hand calculations. See link to LTSPICE from Analog Devices (


Before we look at the specific Dynaco PAT-5’s voltage regulator and low-level preamp circuit, we should examine the emitter follower circuit. See Figure 10-4.

FIGURE 10-4 An emitter follower circuit on the left side, and on the right side a block diagram model of the emitter follower circuit for AC signals.

An emitter follower circuit is often used as a gain of 1 buffer amplifier to provide a high-resistance input (Rin1) and a low-resistance output (Rout_E1).

The input resistance Rin1 ~ β re 1 (β 1 1)(R1 || RL1), where re 5 0.026 v/IC wherein IC is the DC collector current.

And the output resistance from Q1’s emitter terminal before being connected to R1 and RL1 is Rout_E1 5 [Rs1/(β 1 1)] 1 re.

For example, suppose Vin is an audio signal generator with a 600Ω as its

source resistance, Rs1 600Ω, and R1 1130Ω so that IC 10 mA. Then re 5 0.026 v/0.010 A or r2.6Ω. Let Q1 5 PN2222 (TO-92 EBC pin out) with β 5 75, then we have the following with R1 || RL1 530Ω given RL1 1000Ω.

Rin1 ~ β re 1 (β 1 1)(R1 || RL1) 5 75 326Ω 1 (75 1 1)(1130Ω || 1000Ω) Rin1 ~ 1950Ω 1 40.28KΩ or Rin1 42.23KΩ

Rout_E1 5 [Rs1/(β 1 1)] 1 re 5 600Ω/(75 1 1) 1 2.6Ω 5 7.89Ω 1 2.6Ω or

Rout_E1 5 10.49Ω

As we can see, indeed the input resistance, Rin1 5 42.23KΩ, is reasonably high and the output resistance, Rout_E1 5 10.49Ω, is fairly low.

One other characteristic of the emitter follower is that it has power supply rejec- tion. That is, if there is noise or ripple from the power supplies, little of it shows up at the output, Vout_E1. For example, if there is even large amounts of noise (e.g., 1-volt peak to peak ripple) on the 112-volt supply in Figure 10-4, the emitter fol- lower will reject essentially all of the positive supply ripple voltage at Vout_E1. If

there is ripple on the negative supply (–12 v), it will be reduced via a voltage divider as seen from the –12-volt supply via R1 and Rout_E1. The resulting voltage dividing factor is Rout_E1/(Rout_E1 R1), which is 10.49/(10.49 1 1130) ~ 0.01 ~ 1%. For example, if there is 100 mV of ripple on the negative supply for this example, then Vout_E1 will have only about 1 percent of 100 mV or 1 mV of noise at the output of the emitter follower.

NOTE: C1 has large capacitance to provide an AC short circuit (e.g., C1 is ~ 0Ω for AC signals when compared to RL1’s resistance value).

Figure 10-4 shows an example of this circuit with an extra resistor, R2, which does not affect Rin1 and Rout_E1. Collector series resistor R2 can be used as a cur- rent-limiting resistor to prevent Q1 from sourcing excessive current that may dam- age it. Typically, R2 has a resistance value in the order of 2.2Ω to about 100Ω to provide current limiting just in case the Q1’s emitter is shorted to ground. Also, R2 can be chosen with generally a higher-resistance value such as 510Ω to 10KΩ to act as a load for a common emitter amplifier for Q1. Care must be taken such that there is sufficient DC voltage across the collector-emitter terminals to avoid having Q1 in the saturation region, which will cause Q1 to not amplify.

Dynaco PAT-5 Low-Level Preamp Section and Power Supply

In the mid-1970s, the Dynaco corporation designed one of their last high-fidelity preamps, the PAT-5. We will look at a couple of its circuits for estimating perfor- mance and troubleshooting. See Figures 10-5 to 10-7. The design included a phono preamp section via an equalization network and a microphone channel. These fig- ures have had their original Dynaco component reference designations renumbered for this book.

Let’s split Figure 10-5 into two parts—the power supply–voltage regulator and the low-level preamp section. See Figure 10-6 for the power supply–regulator section.

The power supply uses a 90-volt RMS center tapped (CT) transformer, T1, that delivers 45 VAC RMS into each diode’s anode for D1 and D2. The peak DC voltage into C1 is then 1.414 3 45 volts ~ 65 volts peak DC voltage. The cathodes of D1 and D1 provide an unregulated 1 65 volts DC into capacitor C1 via the full-wave recti- fier circuit of D1 and D2. R1 and C2 form a low-pass filter that removes most of the ripple (e.g., a 120 Hz sawtooth waveform for the 60 Hz transformer T1). A quick approximation for the cut-off frequency requires having R2 in parallel with R1 because ZD1, a 43-volt Zener diode, is turned on and for practical purposes the ripple voltage at ZD1’s cathode is nonexistent. Thus, ZD1’s cathode looks like an AC ground for R2 connected to C3. To find the cut-off frequency in a low-pass filter, we have to look at not only the driving resistor into C2 but also any resistor connected to C2 whose other end goes to AC ground. When we have this, the actual RC

Tekstvak: 222

FIGURE 10-5  Dynaco PAT-5 low-level preamp section and power supply with regulator.

Tekstvak: 223

FIGURE 10-6  Power supply and voltage regulator section of the Dynaco Pat-5 preamp.

224    C h a p t e r Te n

time-constant of the circuit is τ 5 (R1 || R2) C2 and the cut-off frequency fc 5 1/[2 π τ] 5 1/[2 π (R1 || R2) C2] 5 1/[2 π (1100Ω) 220 μf] or fc 5 0.66 Hz. For a rough estimate of how much the 120 Hz is attenuated we just take the ratio of the two frequencies, fc / 120 Hz 5 0.66 Hz/120 Hz 5 0.0055 or 0.55 percent of the ripple voltage from the 165-volt unregulated supply. For example, if there is a 1-volt peak to peak 120 Hz sawtooth waveform at 165 volts, then we can expect about 0.55 percent of 1 volt or about 0.55 mV of ripple on the 1 terminal of C2. Because R1 and R2 have equal resistances, they form an averaging voltage summing circuit. Whenever you have two equal resistors and there are two different voltages, V1 and V2, driving each resistor, the voltage at the junction of the two resistors are (V1 1 V2)/2. We have 65 volts at C1 and R1, and 43 volts at R2 and C3. Thus, the estimated voltage at C2 is (65 v 1 43 v)/2 or 1 54 volts DC at C2’s positive terminal. Put it another way, the voltage at C2 should be the halfway point of the two voltages on each side of C1 and C3.

Transistor Q3 forms an emitter follower circuit with 43 volts at its base terminal from ZD1. We will lose about 0.7 volts due to the VBE or base emitter turn-on volt- age. So, at Q1’s emitter we would expect something like 42.3 volts. Q1’s emitter can then provide about 42.3 volts at high currents while allowing ZD1 to be turned on. For example, the Zener diode current is approximately the voltage between R1 and R2 divided by (R1 1 R2). This current is then (65 v – 43 v)/(R1 1 R2) 5 22 v/(2200Ω 1 2200Ω) or ZD1’s current 5 5 mA. If we have 1 mA of base current from Q3, the Zener current will be reduced to 4 mA, which is still good for turning on ZD1. However, with 1 mA of drive into the base of Q3, and with Q3 having a β ≥ 50, at least 50 mA can be provided via Q3’s emitter output terminal to the preamp circuits.

Capacitor C4 (0.1 μf) does not reduce any residual 120 Hz ripple voltage because the capacitance is too small. However, it does get rid of any higher-frequency noise, and C4 also “stomps” out any chance for Q3 to parasitically oscillate at very high frequencies (e.g., > 10 MHz). Further low-frequency filtering is provided by R3, which is coupled to C5, 100 μf at the preamp. (See Figure 10-5.)

Troubleshooting the Power Supply and Regulator

1.            Confirm that the Zener diode, diodes, transistor, and capacitors are installed correctly. For example, if a 43-volt Zener diode is put in backwards, you will get a “fake” Zener voltage of 0.7 volts instead of the correct 43 volts. Remember in a Zener diode, the positive voltage via a resistor or current source will flows into the cathode. A transistor installed backwards, such as having its collector and emitter terminals reversed, may turn into a 7-volt Zener diode and it may have excessive base current due to running the transistor in reverse β mode. For example, the reverse β mode or reverse current gain mode in a transistor results in the current gain dropping to about 1 to 5, whereas a transistor normally has a current gain of over 30.

2.            For your safety, be sure to avoid areas where the AC power line voltage comes in such as the power switch and or the primary windings of T1, the power transformer.

3.            With the power turned on, confirm with a volt meter (negative or black lead at ground) that on the anode of D1 and D2 that there is about 40 AC volts RMS to 50 AC volts RMS. Most DVMs will measure AC in RMS voltages.

4.            Then with the volt meter set to DC measurements, confirm that there is about 65 volts DC on C1’s positive terminal. If the DC voltage is low, the capacitor may be old and have lost capacitance. Parallel a ≥ 220 μf 100-volt (extra) external capacitor with C1 and see if the DC voltage rises to about 65 volts. If the voltage is restored to about 65 volts, replace C1 with the external capacitor. Note that C1 can actually be 220 μf to 470 μf at 100 volts or more. Be careful to discharge the extra external capacitor by connecting across a 47KΩ to 220KΩ resistor for about a minute. If you short out the extra capacitor to discharge it, you will encounter a large spark, which may be unsafe.

5.            Measure the DC voltage at C2’s positive terminal. It should be higher than 43 volts (e.g., ≥ 50 volts DC) if there is about 65 volts at C1’s positive terminal. Then confirm that there is about 43 volts at C3, anode of ZD1, and the base of Q3. If this is confirmed, Q3’s emitter should have about 42.3 volts. You should also confirm that Q3 has a base-to-emitter DC voltage of about 0.7 volt. If the voltages are lower than expected at the base-to-emitter of Q3, then confirm there is about 65 volts at the collector. If Q3 does not have collector voltage > 43 volts, then the base-emitter junction acts like a diode, and excessive base current will be drawn because Q3 is not acting like an emitter follower circuit having a high-input resistance at the base. Also, if the base-to-emitter voltage is greater than 0.7 volts, such as VBE > 1 volt, then Q1’s base-to-emitter junction may be blown, which results in an open circuit. You will need to replace Q3 and a suitable NTE replacement part is found via index.php.

6.            If the voltage at Q1’s emitter is too high, that is, greater than the Zener voltage, 43 volts, then there could be a short across the collector-to-emitter junction of Q3. You will need to replace Q3.

7.            Generally, if the Zener diode is damaged, it can be shorted out, which can result in 0 volts at the base and emitter of Q3. If the Zener diode is an open circuit, Q1’s base voltage will be higher than 43 volts.

8.            Measure the DC voltage across R3 (470Ω), which should be a few volts DC at most. If the voltage across R3 has over 6 volts, then the preamp may be drawing excessive current via C5 or other circuits.

Now let’s look at the preamp section in Figure 10-7.

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FIGURE 10-7  Preamp section of a modified PAT-5 preamp.

This preamp has two feedback loops. One is a DC bias feedback loop that includes R15, R11, R9, and R8. The second is an AC signal feedback loop via R19 and R10 when Switch 1 is at position “3” (microphone), and also with C10, C11, R16, R17, and R10 when Switch 1 is at position “1” (phono cartridge).

Preamp’s DC Bias Point Estimates

We will look at the DC feedback loop first. Q1 by itself is a common emitter amplifier, which means the collector of Q1 has an inverted phase relative to the input signal into its base. In Figure 8-19 of Chapter 8 we saw a common emitter amplifier with a feed- back resistor (R_BC) from collector to base that sets up the collector current of the transistor (along with base to ground resistor R_BE in Figure 8-19). When we look at Figure 10-7 we notice that Q1’s collector is coupled indirectly back to its Q1 base by a “circuitous” route through Q2. This route starts from Q1’s collector connected to the base of Q2, and then via the emitter of Q2 by way of R14 and R15, a feedback network of R11 and R9 feeds a DC signal back to the base of Q1 via R8.

In terms of DC biasing, Q2 is an emitter follower amplifier for Q1’s collector output. Since an emitter follower has no phase inversion from base to emitter of Q2, we can connect a feedback resistor R11 back to the base of Q1 (via R8). Resistor R9 forms a voltage divider with feedback resistor R11. Notice that R11 is connected to C8, a large capacitance capacitor to “completely” filter out any AC signal via Vin such that only a DC signal voltage is fed to R11 to work properly for the DC feedback loop. In Figure 10-6 we can estimate VCC ~ 40 volts for a few volts drop across R3. Let’s look at estimating Q1’s collector current. It is important to note that Q1’s collector (VC1) is connected to Q2’s base. We first want Q1’s collector voltage, VC1, to be much less than one-half of VCC ~ 140 volts. The reason for this is to maximize voltage swing at Q2’s collector output. Ideally, we want the base voltage of Q2 to be sitting at a low voltage such as less than 6 volts, but high enough to bias Q2’s base- emitter junction and provide sufficient voltage at R15 to bias Q1’s circuitry via base resistor R8. An example of a “bad” design would have the collector DC voltage at Q1 at +35 volts. Given that we have a 43-volt supply, this would mean Q2’s collector can swing only from 35 volts to 43 volts, which is about 8 volts peak to peak. However, a better design having Q1’s DC collector voltage at 6 volts instead means that Q2’s output can swing from 6 volts to about 43 volt or about 37 volts peak to peak, which

is much better than having an 8-volt peak to peak swing.

Let’s assume that VC1 is in the order of 5 volts to 10 volts so that the DC voltage at the base of Q2 will allow good output voltage swing at Q2’s collector. Recall that the collector voltage should always be above the base voltage. The collector current ICQ1 5 (VCC – VC1)/R12. If we pick a guess of VC1 5 6 volts, then:

ICQ1 5 (40 v – 6 v)/27KΩ or ICQ1 ~ 1.26 mA

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NOTE: If we had guessed that VC1 = 10 volts instead, then ICQ1 ~ 1.11 mA, which is not much of a difference from 1.26 mA.

Now we can estimate Q1’s DC emitter voltage as:

VE1 5 ICQ1 3 R10 ~ 1.26 mA 3 470Ω or VE1 5 10.59 v DC

We can estimate that the VBE turn on voltage of Q1 and Q2 is about 0.7 volt.

Because the DC current gain for Q1 is very high, we can “ignore” base currents flowing through R8, so there is almost no voltage drop across R8. Thus, Vb1 ~ Q1’s base voltage. In practice if Q1’s β ~ 400 and Q1’s collector current ~ 1.26 mA, its base current is about 1.26 mA/β or about 3.15 μA. The voltage drop across R8 is then 100kΩ 3 3.15 μA or about 0.315 volt.

Q1’s DC base volts ~ Vb1 5 VBEQ1 1 VE1 ~ 0.7 v 1 0.59 v or Vb1 ~ 1.29 volts. We can now find V_R15 via a voltage divider formula:

V_15 [R9/(R11 1 R9)] 5 Vb1

Or V_R15 5 Vb1/{[R9/(R11 1 R9)]} 5 Vb1[(R11 1 R9)/R9]

V_R15 5 Vb1 [ (R11/R9) 1 1]

With Vb1 5 1.29 volt, R9 5 10KΩ, and R11 5 18KΩ, we have for V_R15: V_R15 5 1.29 v [ (18K/10K) 1 1] 5 1.29 v [1.8 1 1] 5 1.29 v [2.8]

or V_R15 5 3.6 volts

Thus, ICQ2 5 V_R15/R15 5 3.6 volts/2200Ω or ICQ2 5 1.64 mA.

And this means VC2 5 VCC – ICQ2 3 R13 ~ (40 v – 1.64 mA 3 10KΩ) VC2 5 (40 v– 16.4 v) or VC2 5 25.6 volts DC.

The estimates for the bias points can be given a ± 20 percent tolerance. We are not after the exact values but something within the “ballpark.” In some cases, you can find the schematic or service manual to high-fidelity product on the web. One such site,, requires login information including email address and password. See Table 10-1 from the calculated values versus information from the Dynaco PAT-5 Preamplifier’s construction and service manual.

TABLE 10-1 Estimated Calculations Versus Service Manual’s Information, Which Are Within About 20 Percent

Voltage NodeEstimated CalculationService Manual’s Information
Q1 emitter voltage (VE1)0.59 volt0.65 volt
Q1 base voltage (Vb1)1.29 volts1.25 volts
Q1 collector voltage (VC1)6.0 volts5.0 volts
Q2 emitter voltage (V_R15)3.6 volts4.4 volts
Q2 base voltage (VC1)6.0 volts5.0 volts
Q2 collector voltage (VC2)25 volts20 volts

Making estimates of the DC bias points can be handy because the service manual or schematic may not always be available. You may have to “manually” draw out a schematic by tracing the circuit you are troubleshooting. From that, many discrete amplifiers can be analyzed quickly either by making estimates as shown here, or by entering the schematic into a simulation program such as SPICE. You can download a free version from Analog Devices via the following link: en/design-center/design-tools-and-calculators/ltspice-simulator.html.

AC Analysis

For looking into the small signal gain of Figure 10-7, we notice that there is an AC feedback path via Q2’s collector to R10 at the Q1’s emitter either through C10, C11, R16, R17, or through R19. In the larger scheme of things, Q1’s base is like the non-inverting input terminal of an op amp or feedback amplifier, while Q1’s emitter terminal is like the inverting input terminal. In the simplest case where Switch 1 selects position 3 (for microphones) to utilize R19 and R10 as the feedback elements, the gain of the preamplifier is Vout/Vin ~ [1 1 (R19/R10)]. If this equation looks familiar, it is the same equation for calculating voltage gain for an op amp circuit for a non-inverting amplifier. With R19 5 68KΩ and R10 5 470Ω, Vout/Vin ~ [1 1 (68K/470)] or Vout/Vin ~ 145.

With Switch 1 set for phono preamp, the feedback network C10, C11, R16, and R17 works like a frequency dependent “resistance.” This feedback network complies to the RIAA phono equalization curve. This network is more properly termed as an impedance network. At high audio frequencies the two capacitors will dominate the impedance characteristic and act like a low-value impedance. We can model this network to give an idea of what we would expect the gain to be as a function of fre- quency. Also, with a signal generator and oscilloscope you can confirm the following:

Vout/Vin ~ [1 1 (Znetwork/R10)]

where Znetwork is a combination of C10, C11, R16, and R17.

At a low frequency such as 20Hz, the capacitors have very high impedances and

the resistors dominate so Znetwork ~ R16 1 R17 5 27KΩ 1 470KΩ 5 497KΩ. Thus, at around 20 Hz, the expected gain within about 20 percent is Vout/Vin ~ [1 1 (497K/470)] ~ 1000. Try setting the generator at 10 mV peak to peak at 20 Hz for Vin, and Vout should be about 10 volts peak to peak within 20 percent.

To find what we may expect at 1000 Hz, we can look at the RIAA feedback net- work and pick out R16 at 27KΩ as the resistor value we will use to find the gain. This means at 1000 Hz the gain will be [1 1 (27K/470)] 5 58.4 within 20 percent. The actual gain may be slightly higher, like 70. Again, you can set the generator at 1000 Hz with 50 mV peak to peak at Vin and Vout should be about 2.9 volts to 3.5 volts peak to peak.

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Finally, at 20 kHz, the RIAA equalization curve requires that the gain is dropped to 10 percent of the 1000 Hz gain. So, at 20 kHz we will expect Vout/Vin ~ 5.84 within 20 percent. Again, set the generator to about 100 mV peak to peak at 20 kHz for Vin, and Vout should be about 0.6 volt peak to peak within 20 percent. If Vout is around 0.75 volts peak to peak, that’s fine too.

Another way to test this is: Set the generator to about 10 mV RMS or about 28 mV peak to peak at Vin, and test at 20 Hz, 1000 Hz, and 20 kHz and note Vout’s amplitude. At 20 Hz, the amplitude should be about 10x of the output at 1000 Hz. And at 20 kHz, its amplitude should be about 10 percent or one-tenth of the ampli- tude of the 1000 Hz signal.

Troubleshooting the Preamp in Figure 10-7

•     Make sure there is about 140 volts DC from the power supply to C5 and then check for the DC bias points listed on Table 10-1. If the DC voltages are within about 20 percent, then the preamp is ready to be tested with AC signals.

•     If there are DC bias problems, with a DVM, check the VBE (base to emitter) turn-on voltage for transistors Q1 and Q2. Typically, this should be between

0.6 volt and 0.7 volt. If you still have problems and the DC voltages are incorrect per Table 10-1, inspect each of the transistors to make sure the pin outs for emitter, base, and collector are connected correctly. For example, if any of the transistors are connected in reverse, where the collector and emitter terminals are reversed, the β will drop from about 400 or 500 to 1 to 5, which will cause excessive base currents to be drawn. If there is excessive base current from Q1, the DC voltage across resistor R8 will be large, since typically with a β ≥ 400, the voltage across R8 will be < 0.5 volt. If Q1 is wired with collector and emitter reversed, the voltage across R8 will be >> 0.5 volt.

•     Check that all electrolytic capacitors are installed correctly polarity-wise. An incorrect installation in electrolytic capacitors will cause capacitor leakage currents, which can “throw off ” the DC bias voltages.

•     With a signal generator at 1000 Hz sine wave and 50 mV peak to peak for Vin, select Switch 1 for phono operation, and confirm that Vout has about 60 to 75 times the input amplitude or 3 volts to 3.75 volts peak to peak. You can probe Q2’s base to verify that you get about 100 times less amplitude of the 1000 Hz signal than at the collector. The gain of the second stage Q2 is ~ R13/R14 5 10K/100 ~ 100 within 25 percent. So expect Q2’s base to have about 30 mV of signal. Note that R15 is AC shorted to ground via C8. Confirm that there is no 1000 Hz sinewave signal C8 (220 μf) with an oscilloscope set at 10 mV per division sensitivity and 500 μsec per division. You should get close to a flat line at this oscilloscope setting. If there’s an AC signal at C8’s plus terminal, replace it with a new 220 μf capacitor of at least

50 working volts. Also confirm that Q1’s emitter terminal has almost the same AC amplitude voltage as Vin. Note that Q1’s base and emitter terminals are like the (1) and (–) inputs to an op amp, which have the same AC voltage (e.g., virtual short circuit across the inputs) when negative feedback is applied.

•     If Vout seems to have some high-frequency parasitic oscillation, check or replace C7 (47 pf) since this capacitor is used to ensure that this preamplifier is stable with feedback components ( R16, R17, C10, and C11).

•     If the Vout’s signal is clipped, check the load resistance value, which typically should be ≥ 10KΩ. A typical audio power amplifier has at input load resistance in the range of 10kΩ to 250kΩ. If the preamp’s output is loading into a headphone or resistor in the 1KΩ range, then Vout will clip at voltages

< 3 volts peak to peak. Again, this preamp is not meant to drive earphones or headphones. The Q2 output stage can only deliver about a milliamp or two.

A High-Fidelity Audio Power Amplifier

Now let’s look at a SX-636 Pioneer stereo receiver power amplifier. Although this is a 1970s amplifier design, the circuit topology still held up into the early 2000s. Many of the newer designs still include its basic features, a differential pair input stage and a complementary push pull output stage. We will explore some of the sub-circuits in this amplifier including the input stage, the second voltage gain stage, and its output stage.

As engineers learned how to design with transistors, they migrated from earlier circuit topologies that included audio signal transformers to output transformer-less designs using output capacitors (e.g., 2200 μf) to direct DC coupled output stages. The SX-636 stereo receiver’s design used positive and negative power supplies, and in essence, it is almost like a high-power output op amp. See Figure 10-8.

Most power amplifiers include two-stage preamplifiers, and this circuit is no exception. In place of a single transistor difference amplifier from Figure 10-7 Q1 (BC414C) where Q1’s base is the (1) input terminal and its emitter is the (–) input terminal, we have in Figure 10-9 transistors Q1 and Q3 where the Q1’s base is the

(1) input terminal and Q3’s base is the (–) input terminal.

In Figure 10-9, Q1 and Q3 form a “true” differential amplifier where the input resistance input into each base is the same. Recall that in a single transistor difference amplifier, the input resistance into the base is much greater than the input resistance into its emitter by a factor of β. Note: Start with confirming all supply voltages.

We can observe that the emitter-to-base voltages of Q1 and Q3 are both approx- imately 0.7 volts. From this we can also surmise that the DC voltage across VB1 and VB3 is ~ 0 volt since VB1 5 VBEQ1 1 VEBQ3 1 VB3, or VB1 – VB3 5 VBEQ1 1 VEBQ3.

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FIGURE 10-8 A redrawn full-page schematic of the Pioneer SX-636 audio power amplifier. Sections A = first stage differential pair, B = second stage voltage gain amplifier, and C = output stage to supply large currents to a loudspeaker.

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FIGURE 10-9  SX-636 power amplifier schematic.

However, VBEQ1 5 –VEBQ1, thus VB1 – VB3 5 (–VEBQ1 1 VEBQ3). Since

the emitter base turn-on voltages of PNP transistors Q1 and Q3 are both ~ 0.7 volt, so VEBQ1 5 VEBQ3 5 0.7 v. And this means the difference voltage between the two bases is VB1 – VB3 5 –0.7 v 1 0.7 v, or VB1 – VB3 0 v, which also leads to VB1 VB3, which is the same approximation for an operational amplifier circuit with negative feedback where the (1) and (–) inputs are about the same voltage. The output of the Q1 and Q3 amplifier is VC1 that feeds the amplified signal to the sec- ond stage, Q5, a common emitter amplifier. The output signal from Q5’s collector (VC5) is then fed to a two-stage complementary transistor emitter follower with the first emitter follower stage formed by Q7 and Q9, and whose second stage emitter follower circuit includes Q11 and Q13, which are high-current output transistors that are mounted on a heat sink. Q7, Q11, Q13, and Q9 biasing diodes are D5A-D5C and D3. These diodes are also on the heat sink with output transistors Q11 and Q13. Diodes D5A-D5C and D3 then track the temperature of the output devices and pre- vent Q11 and Q13 from heat damage. For example, if the output transistors heat up, diodes D5A-D5C and D3 will also heat up and provide a lower biasing voltage to the output transistor, therefore reducing the output transistors’ DC quiescent collector current and thus reducing power dissipated (e.g., which lowers the temperature) in the output devices Q11 and Q13.

DC Biasing Conditions in Figure 10-9

Q1 and Q3 form a differential pair amplifier where it is the difference in voltage between their two bases that provides an amplified output at Q1’s collector, VC1. To bias the differential amplifier, resistor R7 provides an emitter tail current to Q1’s and Q3’s emitter. Ideally, when with no signal, Vin  0 volt AC, we want the tail current I_R7 to split evenly between Q1 and Q3. Reminder: Confirm all supply voltages.

When Vin  0 volt AC, Q1’s base voltage, VB1 ~ 0 volt DC. If VB1 ~ 0 volt, then VE1 ~ 0.7 volt since in a PNP transistor the turn-on voltage is 0.7 volt from emitter to base (VEB ~ 0.7 volt).

The emitter tail current is then the current flowing through R7, I_R7 5 [(30 v – 0.7v)/R7] 5 I_R7. With R7 5 100KΩ, we have I_R7 5 (29.3 v)/100KΩ or I_R7 5 0.293 mA. With the tail current 0.293 mA, we have split evenly this current between the collector currents of Q1 and Q3. This means that Q1’s collector current IC1 is one-half of the emitter tail current I_R7, so IC1 5 0.1465 mA.

With Q1’s collector current known, we can take a look at the second stage, Q5’s DC biasing conditions. Q5’s collector needs to draw enough current via IC5 such that at the base of Q7, VB7, we have about 11.4 volts, which is the sum of the VBE turn-on voltages for Q7 and Q11. The reason for having this 11.4 v 5 VB7 is so that by the time we measure Vout, we should get close to 0 volts DC. For example, a large DC voltage of 10 volts DC at Vout can burn out an 8Ω loudspeaker with about 1.25 amps DC flowing through the speaker’s voice coil.

Q5’s collector current, IC5 5 (30 v – 1.4 v)/(R15 1 R17). With R15 5 R17 5 3300Ω, IC5 5 28.6 v/(3300Ω 1 3300Ω) or IC5 5 4.33 mA. Now that we know that IC5 ~ 4.33 mA, we would deduce that the voltage across R19, V_R19, which is Q5’s emitter resistor, should be IC5 3 R19 or about 0.00433 A 3 100Ω 5 0.433 volt across R19, a 100Ω resistor.

The total voltage across Q5’s base emitter junction and R19 5 VBEQ5 1 V_R19

~ 0.7 v 1 0.433 v 5 1.133 volts. This total voltage having VBEQ5 V_R19 1.133 volts is also equal to the voltage across R9, V_R9 plus the voltage across D1, V_D1.

Because IC1 ~ 0.1465 mA, the forward voltage across D1, V_D1 ~ 0.4 v. You can download the spec sheet for the 1S2473 diode for D1.

This means V_R9 1 V_D1 5 VBEQ5 1 V_R19 5 1.133 volts.

Or V_R9 5 1.133 v – V_D1 5 1.133 v – 0.4 v, or V_R9 ~ 0.733 volt.

So, does IC1 3 R9 5 0.1465 mA 3 5100Ω 5 0.733 volt? Let’s see.

0.1465 mA 3 5100Ω 5 0.747 volt

which is pretty close to 0.733 volt.

So, based on working “backwards” from Q5’s collector current for 0 volts at Vout, we see that the voltage driving into the base of Q5 via Q1’s collector voltage VC1 is close but not “exact.” As we will see, this amplifier is part of a negative feed- back system that will adjust VC1 “precisely” to the correct DC voltage such that Vout 5 0 volt DC.

So, how does the amplifier circuit “guarantee” that Vout will have 0 volts DC when Vin 5 0 volt? What if the transistors, Q5, Q7, and Q11, really do not have 0.7- volt VBE base-to-emitter voltages? Wouldn’t we have a DC offset problem at Vout? Fortunately, the amplifier is set up as a unity gain feedback amplifier or voltage fol- lower for DC signals. That is, for DC voltages, you can remove C7 for DC analysis. Now the amplifier looks very much like a voltage follower op amp circuit with R13 as the feedback resistor. So, the amplifier achieves near 0-volt DC at its output due to Vout’ being connected back to the inverting input terminal (Q3’s base) via R13.

Since the differential amplifier Q1 and Q3 and second-stage amplifier Q5 pro- vide “high” DC gain, Figure 10-9’s circuit looks like one “big” discrete op amp of sorts. Thus, by negative feedback, the DC output voltage is very close to 0 volt, because the (1) input (Q1’s base) is set to ~ 0-volt DC, which then drives via feed- back resistor R13 to cause the (–) input voltage at Q3’s base to go to                                                                                                                                                  0 volt. Thus, Vout’ and Vout both go to 0-volt DC. (At Vout, inductor L1 5 0Ω for DC currents.) One of the most important DC conditions to look for is the bias voltages for the driver transistors Q7 and Q9 and the output transistors Q11 and Q13. Biasing diodes D5A – D5C in series with another biasing diode D3 should provide about 12.5 volts from the base of Q7 (VB7) to the cathode of D3. VR1 and R14 supply additional voltage in series with the biasing diodes such that you can set the output transistors’ quiescent bias current. To do this, Vin is removed. Then measure with a volt meter across either output transistor emitter resistors R27 or R29 and adjust VR1 for about

0.025 volt to 0.038 volt so that the output devices are idling at 50 mA for “normal biasing” or at 76 mA for lower crossover distortion.

NOTE: The total DC biasing voltage for the driver and output transistors is VBEQ7 + VBEQ11 + VEBQ13 + VEBQ9 ~ 0.7 v + 0.7 v + 0.7 v + 0.7 v = 2.8 volts.

AC Signal Conditions

For signal tracing this amplifier, we will need an oscilloscope and sine wave genera- tor. Also, because this is indeed a power amplifier, an 8Ω load resistor rated at 40 watts or more will be connected to the output, Vout. You can improvise by connect- ing eight 1Ω 5-watt resistors in series or parallel six 47Ω 10-watt resistors. Although 8Ω loads are commonly used for evaluating power amplifiers, you can also use resis- tive loads in the 4Ω to 16Ω range at the appropriate wattage ratings of at least 40 watts. Alternatively, a 100-watt incandescent light bulb can be used and it will have about 10Ω (cold) resistance, but the resistance will go up as the lamp glows brighter.

If the DC conditions are correct in that Vout ~ 0-volt DC, you can first test the amplifier without any load resistor. Set the sine wave generator for 1 volt peak to peak at 400 Hz. The AC gain should be Vout/Vin 5 [1 1 (R13/R11)] or [1 1 (27K/1500)] 5 Vout/Vin 19. This means Vout 5 19 3 1 volt peak to peak or Vout 5 19 volts peak to peak 400 Hz sine wave.

Now with the oscilloscope, you should be able to also measure about 19 volts peak to peak sine wave at VB7, VB9, and at C9’s positive terminal, Vout”.

The gain of the second stage, Q5, may look “obvious” because the collector load resistors, R15, in series with R17 may look like the total load resistance of R15 1 R17 5 3300Ω 1 3300Ω 5 6600Ω. The DC load resistance is 6600Ω for Q5 and the DC gain from VB5 to VC5 ~ 6600Ω/R19 5 6600/100 5 66. However, the AC load resis- tance is much higher due to AC coupling capacitor C9, which is also known as a “bootstrap” capacitor. Capacitor C19 has two functions. First, it provides a same phase signal from Vout back to R17 for providing extra signal during the positive cycle of the sine wave such that there is increased current drive into Q7 and Q11.

Without C9, the amplifier will clip prematurely at the positive cycle of the sine wave. The second function of C9 is to boost the AC load resistance for Q5’s collector. We can estimate the collector load’s AC resistance due to C9 and R17, which will result in something >> R17. For a very rough approximation, the AC signal is identi- cal at both terminals of R17, which are at Vout” and VB7. With “equal” AC voltages at both ends of R17 no AC current flows through R17, and this results in an infinite AC resistance for R17. However, there is a slight loss at Vout and Vout” compared to the AC signal at Q5’s collector. The two-stage emitter followers from Q7 and Q11 form a gain of about 0.8 to 0.9 gain. That is Vout/VB7 ~ 0.8 to 0.9. The bootstrapping effect from C9 that couples the output signal Vout back to R17 will raise the effective

collector load resistance for Q5 in the following manner: Q5’s load resistance ~ R17/[1 – (Vout/VB7)]. So, if Vout/VB7 ~ 0.9, then Q5’s load resistance ~ 3300Ω/[1 – (0.9)] 5 3300Ω/0.1 5 33,000Ω, and the approximate AC gain from Q5 is about 33,000Ω/R19 5 33,000/100 5 330. Note that R19 5 100Ω.

Compared to the DC gain of 66, the AC gain of 330 is 5 times higher due to the bootstrapping capacitor. The actual AC gain from VB5 to VC5 will be at least twice the gain of 66 but likely less than 500 when loaded with an 8Ω load at Vout. If the gain is around 66 or less, then C9, the bootstrapping capacitor, has lost capacitance and needs to be replaced. For example, if the output voltage is 19 volts peak to peak at 400 Hz sine wave, then VC5 should also have about 19 volts peak to peak; and the VB5 should have < (19/66) volt peak to peak, which is < 287 mV volt peak to peak at Q5’s base. Note that in this example, VB5 ~ 19 volts/AC gain, and if the AC gain is estimated to be at 330, then VB5 ~ 19 volts/330, or VB5 ~ 57.5 mV peak to peak (which is less than 287 mV peak to peak).

NOTE: The diodes D5A to D5C and D3 form a low AC resistance ≤ 50Ω because they are being used as a constant voltage source, which by definition has ~ 0Ω AC resistance. The maximum resistance by VR1 in parallel with R14 is about 70Ω.

Troubleshooting the Power Amplifier

1.            With the power turned off, check the resistances from collector to emitter of the output transistors Q11 and Q13. You will need to measure the collector- to-emitter resistances a second time with the probe leads reversed. Typically, if the resistances are > 10Ω, the transistors are fine. But if the collector-to- emitter resistances fall in the 2Ω or less category, chances are that there is a shorted transistor. If you find any one of the output transistors shorted, remove both output transistors Q11 and Q13. See Figure 10-10.

Now, measure the collector-to-emitter resistances of the driver transis- tors, Q7 and Q9, the same way as before, twice with the second time having the ohm meter’s leads swapped. If there are no short circuits from collector to emitter of the driver transistors, you can now power up the amplifier without the output transistors, because the driver transistors will keep the negative feedback loop intact via R23 and R25 connected to feedback resistor R13. Confirm that the DC voltage at Vout is about 0 volts. Connect a 1000Ω load resistor to Vout and connect a signal generator at 1-volt peak to peak sine wave at 400 Hz to Vin. Confirm with an oscilloscope that Vout has a sine wave at about 19 volts peak to peak. If Vout does provide about a 19-volt peak to peak signal into the 1000Ω half-watt (or two 2000Ω quarter-watt resistors in parallel) load resistor, then the amplifier probably works fine up to this point and all you need is to replace the power transistors. If you cannot find the original part numbers, you can generally replace the NPN and PNP TO-3

Tekstvak: 238

FIGURE 10-10  Schematic with output transistors removed, and driver transistors Q7 and Q9 to Vout.

case output transistors with parts from NTE at: nte%5CNTExRefSemiProd.nsf/$all/72BF14B50722ACF285257910007E5A0 D?OpenDocument.

Turn off the power or, better yet, unplug the power cord and replace the output transistors. Keep track of which transistor goes where. The NPN tran- sistor will have its collector connected to the 131-volt supply line, which you can confirm with an ohm meter with the power turned off. The NPN 2SD371 can be replaced by an NTE380, MJ802, or 2N5886, and the PNP 2SB531 can be replaced with an NTE281, MJ4502, or 2N5884.

Alternatively, you can select for TO-3 case NPN and PNP power transis- tors that have voltage breakdown of at least 70 volts and at least a 10A collec- tor current rating. For example, the 70-volt transistors 2N3055G (NPN) and MJ2955 (PNP) may serve as replacements. Although, to be on a safer side, the higher-voltage transistors mentioned above should be used. With the output transistors replaced, turn off the signal generator, and turn on the amplifier and measure for about 0 volts at Vout.

2.            Replace the 1000Ω load resistor with at least a 25-watt 8Ω resistor. Be sure to keep the 8Ω load resistor away from other items because it will be very hot once the amplifier supplies signal to it. Turn on the generator for 1 volt peak to peak into Vin at 400 Hz. You should measure with an oscilloscope about 19 volts peak to peak across the 8Ω load resistor. Sweep the frequency from about 50 Hz to about 10 kHz and confirm about 19 volts peak to peak at the output. Now reduce the signal generator’s input voltage by half to 0.5 volt peak to peak, and sweep the generator from 50 Hz to 20 kHz to observe about 9.5 volts peak to peak at Vout.

CAUTION: Do not try to measure the amplifier at 20 kHz at full power. This test can eventually destroy the output transistors due to excessive heating. At the top end of the audio frequencies, the output power transistors become more lossy, and extra heat builds up, which can damage them.

3.            In some cases, it may be difficult to find exact replacements for all the parts. Generally, it is better to replace “up” with higher power dissipation, voltage, or collector current devices. This amplifier has supply voltages of 131 volts to –41 volts. Thus, to be on the safe side, we can specify the transistors’ voltages as > (131 v – –41 v 5 131 v 141 v 5 72 v) or > 72 volts. Given that the AC power line can increase as much as 10 percent, the transistors’ minimum voltage rating would be about 80 volts. For example, you can replace the input transistors, 2SA726 (BCE pin out) with careful installation using 2N5401 (EBC pin out) because the pins outs are not the same. Similarly, the pre-driver transistor, 2SC1451 (EBC), 150 volts, TO-5 case, may be replaced with TO-5 case transistors 2N3440 (EBC), 250 volts or 2N3439 (EBC) 300 volts. If there is a

240    C h a p t e r Te n

recommended substitution with a lower power dissipation or plastic TO-92 case transistor, it’s better to find another transistor with higher power ratings.

For transistors that begin with 2SA, 2SB, 2SC, and 2SD, sometimes the “2S” is dropped. For example, a 2SD371 may be also named D371. Or a 2SA726 and an A726 transistor are really the same electronic part.

However, it is always best to replace any broken transistor with the origi- nal part or with an NTE version. Replacing with other components can some- times lead to parasitic oscillations. Make sure to retest the amplifier with an oscilloscope probing the output terminal, Vout, and look for any small or large high-frequency signals. If they are there, you need to go back and look for original parts.

Also, sometimes you may be able to find the exact part number replace- ments on eBay, but beware of counterfeit items.

This wraps up Chapter 10. Obviously, there are many more ways to examine both the PAT-5 preamp and the SX-636 power amplifier. Troubleshooting takes experience, but make sure to take safety precautions.

Chapter 11 will look into analog integrated circuits.

            CHAPTER 11


Analog Integrated


We will be exploring various types of circuits and systems that use analog inte- grated circuits as examples of troubleshooting, debugging, and improving. Many of the circuits shown in this chapter will have come from the web (world wide web) or from magazine articles or books. When we build hobby circuits, or fix commer- cially manufactured products, generally one would assume that the circuits are reli- able in terms of performance/design with replaceable parts. From my experience building hobby circuits, working in repair shops or radio stations, and teaching students, I have found that troubleshooting circuits can be broken down into the following categories:

•     There are some circuits that will not work or will not work very well as shown in the schematic. These circuits will require some redesigning.

•     There are some circuits that will work as a “one of a kind” but may not be repeatable due to the design. This type of design sometimes works by luck. For example, a photodiode circuit seems to work if a minimum amount of light falls on it because the op amp may not have the capability to output down to 0 volts but have to start at about 11 volt (e.g., see Figure 11-18).

•     Many circuits published really do work. And these circuits will perform fine as shown in the schematics or instructions. However, sometimes it may be “easy” to mis-wire a circuit or put in the wrong part. Other times, the circuit may work in “normal” consumer everyday use but will break down under 24/7 schedules due to overheating, for example. Using a higher-power device can fix this problem.

We will concentrate on the first two categories—circuits that are redesigned for more reliable operation.


Circuits That Need Fixing or Redesigning

Let’s take a look at a problem I had to solve just recently when a young student showed me a preamp circuit lifted off the web for amplifying electric guitar pickups. See Figure 11-1.

FIGURE 11-1 A preamp meant to have a gain ~ 30 but instead has a gain of 1.

As we can see, Figure 11-1 shows some good points in that the DC bias voltage is correct to provide one-half the DC supply voltage, V0, from R3 and R4. Via V0 connected to R5 and R2, the op amp’s output pin will be biased to one-half the DC supply voltage for maximum AC voltage swing. With a 30KΩ R1 feedback resistor and a 1KΩ R2 input resistor, at first glance, the (inverting) voltage gain should be

–R2/R1 5 –30K/1K 5 –30, correct? No. Because the input signal also is coupled to the non-inverting input pin 3, there is a gain cancellation effect going on. The reason is any input signal at the (1) input of op amp with the same input voltage at R2 results in a non-inverting amplifier’s gain summed with the inverting gain.

Note that the input resistance into the (1) input terminal, V2, is very high so there is “no” current flowing through R5, which means there is no voltage across R5, or R5 may as well be a short circuit or 0Ω wire. The reason is because V0 5 V2, and since V2 at the (1) input terminal acts as a reference voltage for the feedback ampli- fier, the (–) input voltage V1 must follow the same voltage as V2. Thus V1 5 V2, and since V0 5 V2, V0 5 V1. This means the voltage across R2 is 0 voltage and the input resistance to R2 (e.g., at V0) is infinite resistance since 0 volts across R2 implies there is no current flowing through R2.

When the input resistor R2 (left side) is connected to the (1) input (via R5) we have the following:

The non-inverting gain is [1 1 (R1/R2)]; and the inverting gain is –(R1/R2).

The total gain (AC or DC signal) is then [1 1 (R1/R2)] 1 –(R1/R2), which leads to: 1 1 (R1/R2) – (R1/R2) 5 1 5 the total gain 5 non-inverting gain plus inverting gain.

This total gain makes sense in terms of the DC gain as well. Since R3 and R4 form a voltage divider of half the supply voltage at R5 and R2, with a gain of 1, the DC output voltage at pin 1 of the op amp has to be one-half the supply voltage. For example, if 1v 5 9 volts, pin 1 5 9 volts/2 or 4.5 volts DC.

So how do we fix this circuit to have gain? See Figure 11-2, where we add C5.

FIGURE 11-2 By adding C5, we provide an AC short to ground at the non-inverting input pin.

By AC shorting to ground the non-inverting input (pin 3) via adding C5, we now have a “standard” inverting gain amplifier that has a gain of Vout/Vin 5 –R1/R2 5

–30K/1K or Vout/Vin 5 –30. The preamp is restored to a reasonable gain for ampli- fying the guitar pickup’s signal. R1 may be increased further to increase the gain, such as having R1 5 100KΩ for a gain of –100.

If we look at the AC input resistance, Rin, of this preamp in terms of Vin “look- ing” into C2, we see that C2 and C5 are AC short circuits, so we have the following pertaining to Figure 11-2.

R3’s top connection is connected to 1V, an AC ground, so AC-wise R3 is in parallel with R4. Because C5 acts as an AC short to ground, R5, which is connected to C5, is grounded as well. This means the AC input resistance has at least R3 in paral- lel with R4 and R5. Because the (1) input is AC grounded via C5, we know that

the input resistance of an inverting op amp circuit is just the input resistor, which is R2. Thus, the AC input resistance is then R3 in parallel with R4, R5, and R2. Or, Rin 5 R3||R4 || R5 || R2. We can group the equal value resistors together since R3 5 R4 5 100kΩ, and R5 5 R2 5 1kΩ.

Rin 5 (R3||R4) || (R5 || R2)

Since R3||R4 5 100kΩ||100kΩ or R3||R4 5 50KΩ, and R5||R2 5 1kΩ||1kΩ or R5||R2 5 500Ω, Rin 5 50kΩ||500Ω. However, Rin ~ 500Ω, since 50KΩ >> 500Ω.

In this first fix in Figure 11-2, the gain was restored, but R5 added to the loading input source, Vin. We can rescale the values for R5 and C5 to increase the input load resistance at Vin. See Figure 11-3.

FIGURE 11-3 We can scale R5 up 100x and also use a smaller value C5 to provide an AC short circuit to ground.

By rescaling the values of R5 and C5, we mean having the time constant R5 3 C5 of Figure 11-3 equal to or greater than R5 3 C5 of Figure 11-2. The reason for this is so that C5 in both Figures 11-2 and 11-3 will be an AC short circuit to ground rela- tive to R5.

In Figure 11-2, the resistor-capacitor time constant R5 3 C5 5 1KΩ 3 33 μf 5

0.033 second; and in Figure 11-3, R5 3 C5 5 100KΩ 3 1 μf 5 0.10 second > 0.033 second, so we are OK with this rescaling. A larger time constant means we have more filtering from noisy power supplies at 1V. In general, C5 should be at least 0.1 μf. That is, if you scale R5 to 10MΩ and C5 to 0.01 μf, the preamp will work providing the circuit is enclosed in a shielded and ground metal box because pin 3 of the op amp will not have a low enough impedance at 50 Hz or 60 Hz and may pick up hum from power lines.

Given the rescaled values for R5 and C5, the input resistance now is: R3||R4||R5||R2 5 (R3||R4||R5) || R2

With R3 5 R4 5 R5 5 100KΩ, (R3||R4||R5) 5 100KΩ/3 5 33.3KΩ, which is

>> R2 5 1KΩ. Thus, Rin ~ 1KΩ in Figure 11-3 versus Rin 5 500Ω in Figure 11-2. Can we improve this circuit further and why? If the power supply is from a bat-

tery, there will not be a problem in terms of noise induced to the input via R3 and R4. However, if 1V is from a power supply with some noise such as power supply ripple, the power supply noise may be coupled to the output of the op amp. See Figure 11-4 for a further improved circuit to reject power supply noise and maintain maximum input resistance for an inverting gain op amp circuit.

FIGURE 11-4 An improved circuit to further reject power supply noise.

By connecting the DC biasing circuit R3 and R4 with C2, a low-noise DC bias voltage is established at the non-inverting input terminal of the op amp. Since R3 and R4 are equal in value, the DC voltage is one-half 1V. Thus, we get one-half the supply voltage at the (1) input terminal of the op amp and because of R1, the DC bias voltage at the (1) input terminal “forces” the same voltage at the (–) input ter- minal. And because there is no DC current flow through R1 (recall input bias cur- rent  0), the voltage across R1 5 0 volts. Thus, the DC output voltage at pin 1 is also one-half the supply voltage, which is (1V/2). Again, DC-wise, R2 is “discon- nected” since C4 is an open circuit at DC.

The values of R3, R4, and C2 were chosen to form a low-pass filter with a very low-frequency cut-off as to substantially reduce any power supply ripple from 1V. With R3 5 R4 5 100KΩ and C2 5 33 μf the cut-off frequency, fcut-off 5 fc:

fc 5 1/[2π(R3||R4)C2] 5 1/[2π(100KΩ||100KΩ)33μf] or fc 5 0.0965 Hz ~ 0.10 Hz

To find the attenuation factor of a noise signal at a frequency fnoise, we have: attenuation factor 5 fc/fnoise 5 0.1 Hz/fnoise

For example, if we have a 60-Hz signal at 10 mV peak to peak riding on top of 1V, then the AC signal at the (1) input will be 10 mV peak to peak 3 (fc/fnoise), which is then 10 mV peak to peak 3 (0.1 Hz/60 Hz) 5 0.0166 mV peak to peak. If more filtering is required, you can increase the values of the resistors R3 and R4 to about 1MΩ with C2 5 33 μf, but it will take about 60 seconds or so for the bias volt- age to come up to one-half 1V.

Note that even though there is the ripple voltage at the op amp’s power pin 8 with 1V, the op amp itself has sufficient power supply ripple rejection to prevent this ripple voltage from feeding through to the op amp’s output pin 1.

If we are concerned with low input resistance, Rin, and power supply noise, one of the best moves is to use a non-inverting gain configuration. See Figure 11-5, which is configured for a higher input resistance at C2.

FIGURE 11-5 A non-inverting gain amplifier with extra power supply filtering via R5 and C5.

By using a non-inverting amplifier for a preamp, we have the advantage of larger input resistance, Rin, which is equal to R4||R3 5 100KΩ||82KΩ or Rin 5 45KΩ, due to C5 being an AC short circuit to ground. Again, if more filtering from noise via the power supply, 1V, is required, C5 can have a larger capacitance such as 330 μf. Generally, a TL082 op amp will be “good” enough low noise-wise for some low-level sources such as magnetic pickups or microphones.

However, you can make a lower noise preamp if the low-level source includes a low impedance device such as a dynamic microphone. In Figure 11-5, you can replace U1A with a lower noise op amp such as an OPA2134, LM833, or NE5532, and rescale the feedback network, R1, R2 for lower resistances such as:

R1 5 3000Ω, R2 5 100Ω, and C4

to a larger capacitance value to maintain good low-frequency response such as C4 5 330 μf.

We now will look at a circuit found off the web or a hobby book. See Figure 11-6, which shows a high-gain op amp circuit that drives a speaker amplifier.

FIGURE 11-6  An amplifier circuit that kind of works, which produces distorted signals.

This amplifier has a basic flaw in that U1 LM741 really requires a negative power supply to pin 4. Also, the other problem with this amplifier is that a DC voltage from the LM741’s output pin 6 couples a DC voltage to the input terminal of U2, LM386. In general, the LM386’s input terminal wants to see an AC coupled signal with no DC component.

As a result of pin 4 grounded for U1, the output signal resembles an amplified half-wave rectifier. See Figure 11-7.

We can show a way to slightly redesign the circuit in Figure 11-6 to solve the problems previously mentioned.

See Figure 11-8 where C5 along with R2 and R3 are added. Please refer to the redesigned circuit in Figure 11-8.

FIGURE 11-7 Top trace is the output signal at about 5.7 volts peak to peak; bottom trace is the input signal at about 124 mV peak to peak.

We can add a minus power supply (e.g., –9 volt) to pin 4, however, a little re- biasing of the LM741’s input stage should fix our problem. See Figure 11-8.

To debug Figure 11-6, we “un-ground” the non-inverting input pin 3 of U1 and provide it a one-half supply voltage at 4.5 volts. C4 is added to reduce noise from the 19-volt supply. Input capacitor C1 is now connected to a series resistor R4 to pro- vide an inverting gain op amp circuit, where the gain of U1 is – R1/R4. If more or less gain is required, the resistance value of R4 may be varied, such as from 1000Ω to 100KΩ. If a better low-frequency response is needed, C1 can be replaced with a 33-μf (25-volt) capacitor with its positive terminal connected to R4 (Figure 11-8).

In Figure 11-6 the signal was coupled directly via C1 to the inverting input ter- minal, which made the op amp work in an “open” loop configuration. When an op amp is operating in an open loop configuration, it is behaving more like a compara- tor and the output signal is more like either on or off. Now see Figure 11-8.

To ensure that no DC voltage appears at the input of U2, an extra coupling capacitor C5 is added in series with POT1. Finally, a snubbing network, R5 and C6, is added at the output of the LM386 pin 5 to ensure that no parasitic high-frequency oscillation occurs as recommended in the LM386’s data sheet.

Other improvements to this circuit can include changing output speaker capaci- tor C3 to a larger capacitance value, such as 100 μf to 220 μf (16 volt).

If a “faster” op amp is needed in terms of gain bandwidth product (GBWP) for U1, the LM741 (1 MHz GBWP) may be replaced with a TL081 (3 MHz GBWP), LF351(4 MHz GBWP), or OPA134 (8 MHz GBWP).

Tekstvak: 249

FIGURE 11-8  A “fixed” version with an extra bias circuit, R2 and R3, which allows the circuit to work off a single supply.

Our next circuit that needs debugging is an active band pass filter that can be used for audio signal applications such as an equalizer, or for applications where you want to pass a signal around a center frequency while attenuating signal whose fre- quencies are outside the center frequency.

For example, if you like to “isolate” a 19-kHz signal that has been added to a high-fidelity audio whose frequency ranges from 50 Hz to 15 kHz, an active band pass filter can be used. It will pass the 19-kHz signal essentially without attenuation, while removing most of the signal from 50 Hz to 15 kHz.

In Figure 11-9 there are two graphs of a lower Q band pass filter and a higher Q band pass filter. Notice that the higher Q filter is narrower in shape, which allows for better rejection of signals whose frequency is outside of the center frequency fc. A good approximation for Q is the center frequency divided by the –3 dB or 70.7 per- cent amplitude frequencies. That is Q 5 fc/BW–3 dB.

FIGURE 11-9  Graphs of band pass filters relating to –3 dB (at 70.7 percent) points and Q.

For the lower Q band pass filter: Q 5 fc/(f4 – f1) where (f4 – f1) 5 –3 dB band- width; and for the higher Q band pass filter: Q 5 fc/(f3 – f2) where (f3 – f2) 5 –3 dB bandwidth.

For example, if the center frequency, fc 5 10 kHz and f3 5 10.5 kHz with f2 5 9.5 kHz, then the higher Q band pass filter has:

Q 5 10 kHz/(10.5 kHz – 9.5 kHz) 5 10 kHz/1kHz or Q 5 10

If the lower Q band pass filter has f4 5 11 kHz and f1 5 9 kHz, and recall that fc 5 10 kHz, then its Q 5 10 kHz/(11 kHz – 9 kHz) 5 10 kHz/2 kHz or Q 5 5.

The graph in Figure 11-9 overlays the “normalized” curves of the lower Q and higher Q band pass filters. In reality, often the higher Q band pass filter will have more gain at the center frequency when compared to the lower Q band pass filter. For example, if the higher Q band pass filter provides 2 volts AC at fc, then the lower Q band pass filter may provide 1 volt AC at fc. Note: On the next page fcenter 5 fc.

Now let’s take a look a band pass filter that does not work with AC signals. See Figure 11-10.

FIGURE 11-10 Example wrong active band pass circuit that has DC biasing problems at the Vin input terminal with R1.

The band pass filter in Figure 11-10 would work if the minus supply pin 11 that is connected to ground would be instead connected to a negative power supply such as

–9 volts. As a result of having the negative supply pin 11 connected to ground, the op amp is not capable of outputting a signal that swings below ground. Also, the non- inverting input at pin 3 cannot tolerate a signal below –0.5 volts. So again, an AC signal connected to Vin will produce a distorted Vout. Also note that the LM324 is known for crossover distortion, so using that op amp is not the best choice along with its low gain bandwidth product of 1 MHz. This bandpass filter can run the limit of an op amp’s gain bandwidth product because it is “gain hungry.” For example, suppose we want to build a 19-kHz bandpass filter with a –3 dB bandwidth of 2 kHz. This means at 1 kHz below and above the center frequency, 19 kHz, the response is 70.7 percent of the peak amplitude at 19 kHz. This means at 18 kHz and 20 kHz these are the –3 dB frequencies. The Q of a bandpass filter is defined as Q 5 fcenter/BW, where

fcenter 5 center frequency of the band pass filter.

With this example, Q 5 19 kHz/2 kHz or Q 5 9.5. The gain at the center frequency is Av 5 2 Q2.

For this example, Av 5 2 (9.5)2 or Av 5 180.

To determine the op amp’s open loop gain at 19 kHz, an approximation is GBWP/ fcenter.

For the LM324 that has a 1 MHz GPWP, the open loop gain at 19 kHz ~

1 MHz/19 kHz ~ 52.

Note that we need at least 180, and generally we want the open loop gain to be about 4 to 10 times of Av 5 2 Q2. That is, we want the op amp’s open loop gain at 19 kHz to be 720 to 1800.

For example, an LT1359 Quad op amp with 25 MHz GBWP would work, having an approximate gain of 25 3 52 5 1300 at 19 kHz.

Sometimes it’s better to use dual op amps rather than quad op amps. The reasons are a wider selection, lower cost, and a greater chance to find a faster op amp. Now let’s take a look at the debugged version in Figure 11-11.

FIGURE 11-11  Corrected band pass filter circuit with a DC bias circuit, R4, R5, and C5.

By adding a bias circuit to form a one-half supply voltage with a large capaci- tance C5, the circuit is debugged. Also, a different quad op amp, the RC4136, which has a 3 MHz GBWP, is used instead of the LM324. The RC4136 has a very good output stage that does not exhibit crossover distortion.

Because the RC4136 has a 3-MHz gain bandwidth product (GBWP), the band pass filter’s center frequency is limited to Q < 10 and fcenter < 3.75 kHz. A worst case calculation was made by the following:

2 Q2 5 2 (10)2 5 2 (100) 5 200

This gain number of 2 Q2 is when R2 is removed from the circuit and R1 drives the band pass filter.

We want an open loop gain of at least 4 times the 2 Q2 at the band pass filter’s center frequency.

The open loop gain required is then 4 3 2 Q2 5 4 3 200 5 800. Now we need to find the band pass filter’s maximum center frequency, fcenter-max:

fcenter-max 5 GBWP/(4 3 2 Q2) 5 3 MHz/800 5 3.75 kHz

To determine the values of the resistors R1, R2, and R3, and the capacitors C1 and 2 for a given band pass filter with a center frequency, fcenter, and Q based on the desired –3 dB bandwidth, see the following:

–3 dB bandwidth 5 fcenter/Q

R1||R2 5 R3/(4 Q2), where also R1||R2 5 (R1 3 R2)/(R1 1 R2) R3 5 Q/(π C fcenter)

C 5 C1 5 C2

The gain of the overall band pass filter is [R2/(R1 1 R2)] 3 2 Q2.

When the value of Q may be in the order of 3 to 10, this gives a gain range, 2 Q2, from 18 to 200. This means that the band pass filter has quite a bit of gain at the center frequency. To ensure that the band pass filter does not overload too easily, the input resistors R1 and R2 form a voltage divider that will reduce the overall gain. A good starting point for R1 and R2 is to make a voltage divider of about 10:1 or 11:1. For example, in an 11:1 voltage divider, if R1 5 10R2, then [R2/(R1 1 R2)] 5 1/11; the gain range from Vin to Vout is now reduced to 1.6 to 18 instead of 18 to 200.

One way to start designing the band pass filter is to determine C as a function of R1||R2, Q, and fcenter. We can substitute R3 5 Q/(π C fcenter) into R1||R2 5 R3/(4 Q2) to express R1||R2 as a function of Q, C, and fcenter, which leads to the following:

R1||R2 5 1/[π C 4Q fcenter]

and now we can find C as a function R1||R2, Q, and fcenter as: C 5 1/[π (R1||R2) 4Q fcenter] 5 C1 5 C2

Once C is determined, we can find R3 via: R3 5 Q/(π C fcenter)

For example: Suppose we want to build a band pass filter having fcenter 5 1 kHz with Q 5 3, and given that R1 5 100KΩ and R2 5 10KΩ such that R1||R2 5 100KΩ||10KΩ ~ 9.1K.

C 5 1/[π (R1||R2) 4Q fcenter] 5 1/[π (9.1KΩ) (4)(3) 1kHz] or C 5 2916 pf 5 C1 5


C1 5 C2 ~ 2700 pf in parallel with 220 pf or a 0.0027 μf in parallel with 220 pf.

R3 5 Q/(π C fcenter) 5 3/(π 2916 pf 1 kHz) or R3 5 109KΩ ~ 110KΩ, and to reiter- ate, R1 5 100KΩ and R2 5 10KΩ.

The overall gain at the center frequency is Vout/Vin 5 [R2/(R1 1 R2)] 3 2 Q2 or in this example:

[10K/(100K 1 10K)] 3 2 (3)2 5 [1/11] 3 2 3 9 5 18/11 5 1.6 5 gain or

Vout/Vin 5 1.6 @ 1kHz.

Photodiode Circuits

We now turn our attention to photodiode preamps. These are often used with sensor or communication systems. Here we will explore some mistakes as posted on the web or in some hobbyist books.

Photodiode sensors with light sources such as LEDs can be used to sense the presence or distance of an object. Also, photodiodes with a focused light beam can be used to read black and white lines such as bar codes. In this section, we will look at photodiode “receiver” circuits that convert the light sensed into a signal voltage. Let’s first look at a typical photodiode that has a cathode lead and an anode lead along with its associated schematic symbol. See Figure 11-12.

FIGURE 11-12 Schematic symbol of a photodiode (note arrows pointing into the diode whereas in an LED the arrows are pointing away from the diode), and identification via the lead length where the anode lead is slightly longer than that cathode’s lead.

If the leads are the same length, you can use an ohm meter to measure the for- ward voltage from anode to cathode, which is about 0.7 volts or so. Measure again with a standard diode (1N914) where the cathode is clearly marked with a band, and you should be able to identify the cathode of the photodiode. Now, let’s take a look at what is really a photodiode. See Figure 11-13.

The photodiode acts like a light intensity current source, with the positive cur- rent flowing from cathode to anode. In a sense it’s like an NPN transistor current source where the current flows from collector to emitter. Hence, the photodiode is

FIGURE 11-13 A photodiode schematic diagram is on the left side, and a current generator representation of it shown on the right hand side with the internal diode, Int_Diode.

polarity dependent. The photodiode current is modeled as “I (brightness)” to depict that it is proportional to light intensity. That is, the more light there is, the higher the current is generated from the photodiode. Typically, the photodiode will generate currents as low as I (brightness) < 0.001 μA when it is covered or in the dark, or with intense lighting such as in sunlight, it will generate I (brightness) >> 1 μA.

In this book, we can also sometimes “shorthand” I (brightness) with I (lum), where “lum” is the luminance level that is the same as brightness. For example, when you buy an LED bulb for a ceiling lamp, it is rated in lumens (e.g., a 9-watt LED lamp equivalent to a 60-watt incandescent bulb gives out about 800 lumens).

Note let’s take a look at some working circuits before we examine non-working ones. A photodiode can be connected to load resistor RL to develop a voltage across it as shown in Figure 11-14.

FIGURE 11-14 Two photodiode circuits, one with a +V bias (left side), and the other on the right side with 0-volt bias (via the cathode being connected to ground).

To “extract” the photodiode’s signal, generally it is reverse biased or zero biased to prevent the photodiode’s current from diverting into its internal diode instead of flowing out into a load resistor. As shown in Figure 11-14, we see on the left side’s circuit that 1V is forming a reverse bias to the internal diode, Int_Diode. The max- imum voltage across RL is VRL 5 (1V) 10. 7 volt.

For example, if we are using a 5-volt supply, then 1V 5 15 volts and the maxi- mum voltage across load resistor RL 515 volts 1 0.7 volt 5 5.7 volts.

The voltage across the load resistor is VRL 5 I (lum) 3 RL and equivalently, I (lum) 5 VRL/RL.

If, for example, RL 5 1MΩ and 1V 5 15 volts, then the maximum photodiode current, I (lum)max 5 5.7 volts/1MΩ or I (lum)max 5 5.7 μA.

In Figure 11-14, the circuit on the right shows what happens when 1V   0 volts

that essentially grounds the cathode of the internal diode. This means that Int_Diode is now in parallel with RL, and also means it “clips” or limits the maximum voltage across RL by the turn-on voltage of a silicon diode or about 0.7 volt. As an example, if RL 5 1 MΩ with 1V 5 0 volts, I (lum)max 5 0.7 volts/1MΩ or I (lum)max 5 0.7 μA

5 700 nA.

If we want to buffer (amplify) the signal output of Figure 11-14, we can use a voltage follower or non-inverting gain amplifier that can sense at ground and pro- vide an output voltage at near ground, such as a TLC272 op amp. However, there can be a disadvantage by merely amplifying VRL with a non-inverting gain amplifier. Because the RL is typically 100kΩ to 1 Meg ohm, even small stray capacitances can cause a frequency response problem.

For example, if the photodiode has a 10-pf internal capacitance, Cpd, across the cathode and anode, the high-frequency response with RL 5 1MΩ will be about 16 kHz where the cut-off frequency, fc 5 1/[2π (RL 3 Cpd)] which is: fc 5 1/[2π (1MΩ 3 10 pf)] 5 15.924 kHz 5 fc.

We can improve on high-frequency response by using a “trans-resistance” ampli-

fier, which means current in and voltage out. But before we show you some trans- resistance amplifier circuits, let’s take a look at testing the photodiode itself with a digital voltmeter (DVM). See Figure 11-15.

Even a typical low-cost DVM can test photodiode via its 200 mV or 2000 mV DC full-scale setting because it has a built-in 1MΩ resistance across the positive and negative leads. You can use a flashlight and shine it directly into the photodiode and read the voltage output with the voltmeter setting. Note that the voltage will be lim- ited to about 700 mV DC. As shown in Figure 11-15, the voltmeter shows 206 mV across the input 1-MΩ input resistance of this particular DVM. This means the pho- todiode is 206 mV/1 MΩ or 206 nA (nano amps) or equivalently 0.206 μA. If you use a higher input resistance DVM that normally has a 10MΩ input resistance at the voltmeter setting, you can measure the photodiode current with 10x sensitivity.

Also shown in Figure 11-15 is that you can measure the photodiode’s current directly by using the 200 μA full scale setting as shown with a flashlight shining directly into the photodiode measuring 54.1 μA.

FIGURE 11-15 Two ways to measure photodiodes via the voltmeter (on the right side) and via the current (micro amp meter) settings (on the left side).

If you need to measure lower photodiode currents, such as using a lower sensi- tivity photodiode or an LED, use a slightly higher-performance DVM with a 10MΩ resistance across the positive and negative leads. With 10MΩ load resistance the sensitivity is increased 10x over the lower priced DVM. For example, most auto- ranging and >$20 DVMs have 10MΩ resistance in the voltmeter mode.

Note if you use an LED as a photodiode, the equivalent internal diode, Int_Diode, as depicted in Figure 11-14 will have a maximum voltage of the LED’s turn-on volt- age, VLED. For example, in a red, green, and yellow standard LED, the turn-on voltage is about 1.7 volts. For a blue or white LED, expect the turn-on voltage to be about 2.5 volts to 3.0 volts. This means if you shine light into the LED, the maximum voltage across RL will be VRL ≤ LED’s turn on voltage, which can be in the order of 1.7 volts or more for VRL depending on what type of LED you are using. Also, the pin out will

be same for the LED as with the photodiode; the shorter lead will be the cathode.

Trans-resistance Amplifiers

Figure 11-16 shows a trans-resistance amplifier with a photodiode connected to the inverting input of an op amp. The input signal is a current and the output provides a voltage. In terms of “gain” we have a voltage output divided by a current input or V/I that has the unit of resistance.

FIGURE 11-16 A trans-resistance amplifier for providing a low-impedance voltage output related to photocurrent from PD1, a photodiode. Note: IR = Infrared and VL = Visible Light.

The photodiode, PD1 in Figure 11-16, has its cathode connected to the (–) input terminal of U1A, which serves as a low resistance point referenced to ground. That is, the input resistance, Rin, looking into the combination of feedback resistor R1 and pin 2, the inverting input of U1A, is very low for two reasons.

The first reason is that since the pin 3 (1) input terminal is grounded (or AC grounded if pin 3 is tied to a 1V positive bias voltage), the (–) input terminal must have the “same” voltage as the (1) input voltage due to negative feedback via R1. Thus, Rin is looking into a virtual short circuit to ground via the (1) input being grounded. In actuality, Vout is adjusting the voltage such that the current through R1 matches the photodiode’s current from PD1. When this happens, the voltage at V(–) is “adjusted” to ground or 0 volts via Vout.

The second reason why Rin is close to 0Ω is that in an inverting gain op amp amplifier, the input resistance looking into the inverting input terminal is Rin ~ R1/[A0(f)], where A0(f) 5 open loop gain of the op amp as a function of fre- quency. You can also estimate A0(f) by knowing the gain bandwidth product (GBWP) for frequencies beyond the open loop frequency response.

For example, if U1A 5 TLC272, the gain bandwidth product, GBWP, is about

1.7 MHz for a 5-volt power supply. If we want to know the approximate Rin at f 5 100Hz with R1 5 1MΩ, we can estimate A0(f) ~ (GBWP/f), with the under- standing that in most cases the maximum A0(f) is between 100,000 and 1,000,000.

Rin 5 R1/ A0(f) ~ R1/(GBWP/f) 5 1MΩ/(1.7MHz/100Hz) 5 1MΩ/17,000 or

Rin ~ 59Ω. When we consider that PD1 outputs something in the order of 10 μA maximum, the voltage at V(–) will be 10 μA x Rin or V(–) ~ 0.59 mV, which is ~ 0 volts.

Note at DC or slowly changing signals with time, A0(f) is ≥ 100,000. So, given that the open loop gain, A0(f) is ≥ 100,000 at DC or low frequencies, Rin ≤ 1MΩ/100,000 or Rin ≤ 10Ω, which pretty much looks like a short circuit to ground considering PD1’s photocurrents are typically ≤ 10 μA.

You can download op amp spec sheets from various manufacturers (e.g., Texas Instruments, Analog Device, New Japan Radio, ST Electronics, etc.) for open loop gain, A0(f), data.

Now let’s take a look at Figure 11-17. The output voltage Vout 5 R1 3 I (lum), where I (lum) is the photodiode’s current flowing downward from cathode to anode of PD1.

FIGURE 11-17 A trans-resistance amplifier with photodiode PD1 in the left circuit with its “Equivalent Model” on the right side showing a light intensity controlled current source generator, I (lum).

Notice in Figure 11-17 the TLC272 op amp is chosen specifically to work with 15 volts for sensing at 0 volts at the input terminals pins 2 and 3, and also for the ability for the output at pin 3 to swing all the way down to ground. Typically, the feedback resistor R1 is in the range of 100KΩ to 4.7MΩ. For example, if R1 5 470KΩ, then Vout 5 I (lum) 3 470KΩ, which is a positive voltage.

Essentially, how we get the positive voltage for Vout goes like this: We know that no current flows into the (–) input at pin 2 of U1, so this means R1’s current IR1 5 I (lum) , the photodiode current. When I (lum) starts generating current based on light intensity, it will “pull” current through R1. Since the (1) input terminal at pin 3 is connected to ground or zero volts, the negative feedback system has to correct itself to ensure the (–) input voltage matches the (1) input voltage, which is ground. Thus, the (–) input is “servoed” to ground via Vout and feedback resistor R1. When the light shines in the photodiode, the photodiode’s current, I(lum), forces Vout, then rises to a positive voltage just enough such that Vout/R1 5 I (lum). This forms a balanced condition whereby the voltage at R1 connected to the (–) input is now

0 volts, which is what the negative feedback system wants. Essentially, it is saying that it wants the following condition at the (–) input terminal Vout – IR1 3 R1 5 0 volts 5 voltage at (–) input, or put in another way, Vout 5 VR1 5 IR1 3 R1.

Since IR1 5 I (lum), then we have Vout – I (lum) 3 R1 5 0 volts at the (–) input terminal.

So Vout is always doing a balancing act to ensure that whatever photodiode cur- rent is pulling from its cathode, the voltage at the (–) input is “servoed” to zero volts. For example, suppose the following in Figure 11-7 (or Figure 11-18 below left side): R1 5 1MΩ and the photodiode current I (lum) 5 0.5 μA. If Vout were to be grounded temporarily, then right side terminal of R1 would be grounded; and the left side of R1 connected to the (–) input terminal would be –0.5 volts which is 5 –I (lum) 3 R1 5 –0.5 μa 3 1MΩ by having the photodiode current “sinking” current instead of sourcing current (e.g., flowing into R1). If Vout rises from 0 volts to 10.5 volts to counter exactly the –0.5 volts, then the voltage at the (–) input terminal will be 0 volts. Thus, Vout 5 I (lum) 3 R1 or in this example

0. 5 μA 3 1MΩ 5 0.5 volt. Note that Vout 5 VR1 1 V(–) or V(–) 5 Vout – VR1, and we want V(–)  0 volts, which leads to 0 5 Vout – VR1 or Vout 5 VR1 that leads to Vout IR1 I (lum), thus Vout 5 I (lum) 3 R1.

Now let’s take a look at some other photodiode amplifier circuits with possible problems and their fixes. See Figure 11-18.

FIGURE 11-18 MC3403P (left side) and LM1458 (right side) photodiode preamps.

In Figure 11-18, the left circuit uses an MC3403P op amp that has an NPN-PNP push pull emitter follower output stage. Because the PNP output transistor’s emitter cannot deliver a voltage to 0 volts, there will be a problem with sensing small photo- currents. For example, if the photodiode current is 0 μA with R2 removed from the preamp circuit, and R1 5 100kΩ, Vout will be in the 150 mV to 1200 mV range when measured with a DVM. To re-establish a truer linear relationship between light intensity and output voltage, pull-down resistor R2 is connected from pin 1 to ground. This ensures that the MC3403P’s output stage works as an NPN emitter fol-

lower capable of operating correctly all the way down to 0 volts. Typically, R2 can be a resistor in the range of 3300Ω to about 1000Ω. Thus, with R2 installed, when the photodiode current is 0 μA, Vout 5 0 volt. Also note that the MC3403P’s input ter- minals will operate properly down to 0 volt.

The LM1458 circuit (right side) in Figure 11-18 will not work because the input circuit of an LM1458 or similar op amp cannot operate correctly when the (–) input terminal is at the same potential as its minus supply pin 4. The (–) input of an LM1458 or similar op amp requires at least 1.5 volts to 2 volts above the negative supply pin (e.g., pin 4).

A fix then is to replace the LM1458 op amp with op amps such as the TLC272 (16-volt max supply), MAX407 (10-volt max supply), NJU7002 (16-volt max sup- ply) or CA3240 (32-volt max supply) whose input terminals can sense at ground and whose outputs can swing to ground.

Suppose we have the correct op amp in terms of input and output voltage ranges, can there be other problems? Yes, see Figure 11-19.

FIGURE 11-19 The photodiode’s internal capacitor, Cpd, can cause oscillations by added extra lagging phase shift in the feedback network via R1 and Cpd.

All semiconductors including diodes and transistors have internal capacitances across their terminals. In Figure 11-19 on the left side, the photodiode, PD1, has a built-in capacitor, Cpd. Typically, Cpd is in the range of 1 pf to as much as 100 pf, depending on the surface area of the photodiode. Since R1 typically has a value in the range of ≥ 100KΩ, the low-pass filter effect via R1 from Vout to V(–) with Cpd adds phase shift that can cause the circuit to oscillate.

For example, if you replace U3A with a faster op amp and or connect Vout to a capacitance load, it is possible to cause the op amp to oscillate. The oscillation fre- quency is generally > 1kHz such as 10kHz or more. To fix this problem, the circuit on the right side of Figure 11-19 shows a feedback capacitor, Cf, in parallel with R1. Cf with R1 forms a positive phase or leading phase network with respect to diode capacitance Cpd.

This positive phase or leading phase offsets/cancels the lagging phase caused by Cpd. At very high frequencies, the impedance of Cf is much smaller than the resis- tance of R1 such that we get a capacitive feedback network of Cf and Cpd. Since a capacitive voltage divider from Vout to V(–) does not exhibit phase shift, the circuit has less chance of oscillation. Generally, we choose Cf for minimum capacitance to maintain good high-frequency response or bandwidth. Typically, Cf is in the order of 1 pf to 10 pf. However, if we need to filter out noise while maintaining a lower but sufficient bandwidth, Cf can be increased.

The –3 dB frequency for bandwidth is: fc 5 1/[2π (R1)(Cf)].

For example, if R1 5 1MΩ and Cf 5 10 pf, then fc 5 1/[2π (1MΩ)(10 pf)] or fc 5 15.9 kHz. If you want twice the bandwidth, reduce the capacitance of Cf to half by setting Cf to 5 pf so that fc 5 31.8 kHz.

We now turn to examples from the web (world wide web) or from some publica-

tions that show mistakes, but these circuits are easily remedied. Let’s start with Figure 11-20.

FIGURE 11-20  The left side circuit shows photodiode PD1 connected in reverse.

In one of the classes I taught concerning photo emitters and sensors, a basic single supply photodiode preamp was built. The output Vout seems to be stuck at 0 volts no matter how much light was shined into the photodiode. As can be seen in Figure 11-20’s left circuit, the photodiode’s anode is connected to the inverting input of U1A. Because the photocurrent positive current is flowing into the inverting input, we can expect to have a negative voltage at the output. This negative voltage output would be true if the negative supply pin 4 of U1 would be connected to a negative supply such as –5 volts DC. However, since pin 4 is grounded at 0 volts, this means the output voltage at pin 1 cannot go lower than 0 volts. Therefore, the output voltage is “clipped” at 0 volts. To fix this problem, just reverse the photodiode’s con- nections as shown in the right side circuit of Figure 11-20. This then has the photo- diode current flowing out from the inverting input that provides a positive Vout. Also, note the op amp that was chosen allows the input terminals to operate cor- rectly at ground, and that allows the output voltage to swing as low as ground. The output voltage for the right side circuit is Vout 5 I (lum) 3 R1. In Star Trek (TOS,

the original series), the science officer usually saves the day by telling the chief engi- neer to invert phase. And this is one time it is correct to invert the phase of the pho- todiode to get the circuit working correctly. Now let’s look at Figure 11-21.

FIGURE 11-21 A example of inverting the phase of the photodiode and the inputs at pins 2 and 3 fixes the problems.

In this example, we have to invert phase twice. See the left circuit in Figure 11-21.

And the mistakes we shall see are somewhat devoid of proper design.

For the left side circuit in Figure 11-21, the photodiode looks similar to the cor- rect orientation for the photodiode’s current flow via I (lum); the configuration, however, is incorrect. Although the cathode of PD1 is properly biased to a positive 15 volts for reverse bias on the photo diode, there are two problems. First there is positive feedback via R1 via the (1) input being connected to the feedback resistor R1. When an op amp is connected in a positive feedback configuration like this, the concept of V(1) 5 V(–) at the input terminals does not apply. Generally, the op amp’s output signal at Vout may show oscillation or may latch to a DC voltage. With positive feedback it does not matter which way the photodiode is connected because the output signal will not be related to photodiode current. To fix the two problems, we have to first, reverse the input pins of the op amp and second, connect the pho- todiode as shown in the right side circuit of Figure11-21. Although the photodiode now sits at 0 volts DC bias, which is fine for most cases, there may be a reason for biasing the cathode of PD1 to a positive voltage. By applying a positive voltage that puts the photodiode into reverse bias, the internal photodiode capacitance reduces. We can take the anode of PD1 in the right circuit of Figure 11-21 and connect it to an available negative power supply. But suppose we only have the positive power supply. Can we still apply a positive reverse bias voltage to the cathode of PD1? The answer is yes. See Figure 11-22, which incorporates a second amplifier (e.g., you can use a dual op amp).

FIGURE 11-22 Biasing the photodiode’s cathode via Vbias and using a differential amplifier circuit via U2B. Note: RR Op Amp = Rail to Rail op amp such as an ISL28218.

We can set a positive voltage on the cathode of PD1 by having the (1) input of U2A biased with a voltage. In this case it is about one-half the supply voltage, or 2.5 volts in this case with a positive 5-volt power supply. Because we have a negative feedback circuit via R1 from Vout1 to the (–) input of U2A, the voltage at pin 2, V(–) is the same as Vbias. That is, under a negative feedback op amp circuit, the voltages at the input terminals are the “same.” Therefore with V(–) 5 Vbias, and Vbias 5 2.5 volts, the cathode of PD1 is sitting at 2.5 volts DC.

The resistors R2 and R3 are chosen such that R2||R3 << R6. In general, we want R2||R3 ≤ 1% of R6. In this circuit, R2 5 R3 5 1KΩ so R2||R3 5 500Ω and 500Ω is

0.5   percent of 100KΩ.

With R2 5 R3, the current draw into R2 is (5 v)/(R2 1 R3) or 5 v/2KΩ 5 2.5 mA. We can increase R2 and R3 to 2KΩ to save power if needed. Decoupling capaci-

tor C3 with R2 and R3 forms a low-pass filter to reduce noise from the 15v supply line. The cut-off frequency of this low-pass filter is fc 5 1/[2π(R2||R3)C3]. In this circuit we have:

fc 5 1/[2π(1000Ω||1000Ω)33μf] 5 1/[2π(500Ω)33μf] or fc 5 9.65 Hz

And generally, we want fc << 60 Hz for good filtering. If needed, we can increase C3 from 33μf to 330 μf. A ≥ 10-volt capacitor should work fine since the supply volt- age is 15 volts and with R2 and R3, nominally, there are only 2.5 volts across C3.

Vout1 is the summation of the bias voltage, Vbias and the photodiode current times R1 or Vout1 5 Vbias 1 I (lum) 3 R1.

Essentially, this circuit is a voltage follower for Vbias since the photodiode, PD1, is a current source that has “infinite” DC resistance. So, there is no DC resistance value from cathode to anode of PD1. Again, see Figure 11-22.

The end game is to provide an output that is in the form of only I (lum) 3 R1, without any DC bias voltage. We then need to subtract out “exactly” Vbias from Vout1 5 Vbias 1 I (lum) 3 R1. By using a differential amplifier U2B, the output of it is Vout2 5 Vout1 – Vbias. Or put in another way (for the circuit in Figure 11-22), Vout2 5 Vbias 1 I (lum) 3 R1 – Vbias 5 I (lum) 3 R1 5 Vout2.

Differential amplifier U2B works as a unity gain subtractor when all four of the resistors, R4, R5, R6, and R7, are the same value. Note that they are 1 percent toler- ance resistors.

Ideally, op amps U2A and U2B are from a dual package rail to rail (RR) output for maximum voltage swing such as NJU7062 or NJM2762. However, if you bias the cathode of PD1 to a slightly lower voltage such as 1.67 volts by having R2 5 2KΩ and R3 5 1KΩ, you can use other op amps such as TLC272 or CA3240 and still get good output voltage swing. Of course, if you just need a slight positive bias voltage at the photodiode cathode, such as a few hundred millivolts, you can set R2 5 10KΩ to have Vbias ~ 10.45 volts.

We now turn to a circuit found in a publication and on the web that is devoid of reliable operation. In a sense it is “bankrupt” in good electrical engineering practice. (After all, this is Chapter 11 and we should cover at least one bankrupted circuit

. . . just kidding!)

See Figure 11-23.

FIGURE 11-23 The left circuit is sometimes called an inverting voltage follower, which is incorrect.

In Figure 11-23, the left side circuit shows positive feedback because the output signal from Vout is connected back to the non-inverting input. This will cause the output signal to oscillate or latch to either a positive or negative voltage. With posi- tive feedback we will find that the voltages measured at the non-inverting and invert- ing input terminals are not equal. That is, for the left circuit with positive feedback V(1) ≠ V(–).

To fix this circuit, you need to reverse the input leads as shown on the right side circuit of Figure 11-23. With this done we now have a standard unity gain non- inverting voltage follower where Vout 5 Vin, and V(1) 5 V(–) .


In this chapter, we have been shown some examples of “bad” circuits, and there is a reason for showing these. As a hobbyist or even an engineer, oftentimes we assume what’s published in books or magazines or posted on the web (world wide web) is correct. In my experience, some of these non-working circuits have “typos,” others include misunderstandings of electronic principles such as the left side circuit in Figure 11-23, and yet other circuits that do not work can be attributed to an incor- rect part or not knowing limitations of components (e.g., rail to rail op amps versus standard op amps). So, when we build such “bad” circuits exactly and check them over and over again and find nothing wrong construction-wise, it’s time to examine the circuit itself, and not assume that it actually works as advertised. As we get out of Chapter 11, which we did in a sense showing how to fix erroneous or “bankrupted” circuits, we will find even more types of these circuits ahead. Stay tuned to the next couple of chapters—more examples to come.

Reference Books

1.            Arthur B. Williams, Electronic Filter Design Handbook, Second Edition.

McGraw-Hill, 1988.

2.            Paul R. Gray and Robert G. Meyer, Analysis and Design of Analog Integrated Circuits, Second Edition. Wiley, 1984.

           CHAPTER 12

Some Ham Radio

Circuits Related

to SDR

There are many amateur radio circuits available today such as those used in software defined radios (SDR). In this chapter we will look into some of these circuits includ- ing building blocks that include crystal oscillator circuits, mixers, and simple RF matching circuits.

We will explore circuits published in magazines and books, and also those posted on the web. Because these are DIY (do-it-yourself) type circuits, there may be room for improving performance and reliability, or room to simplify these circuits.

This chapter will cover mainly amateur radio receiving circuits so that anyone can build and explore these without having to obtain a license (e.g., Technician, General, or Extra).

However, if you are transmitting ham radio signals you will require an FCC (Federal Communications Commission) license in the United States or an equiva- lent amateur radio license elsewhere.

Software Defined Radio Circuits

A software defined radio uses a computer to handle “middle and back” portions (intermediate frequency, IF, filter, demodulator/detector, audio filtering, and audio output circuits) of a superheterodyne (superhet) radio. Essentially, a basic superhet radio includes a tuned circuit for the received signal, a mixer circuit to shift down the received signal’s frequency, an intermediate frequency filter and amplifier, and finally a detector (e.g., AM) and audio amplifier. See Figure 12-1.

NOTE: RF mixing means we are in some way multiplying two signals to get a third signal.


268    C h a p te r T w el v e

FIGURE 12-1 A block diagram of a basic conventional superhet radio.

As shown in Figure 12-1 the basic superhet radio requires an RF band-pass filter (BPF) to ensure that the RF mixer with the local oscillator provides an IF signal indicative of the tuned RF signal from the antenna. For example, in a standard AM radio, the IF is 455 kHz. This means that the local oscillator’s frequency is 455 kHz above the incoming RF signal. But if there is an RF signal that is 455 kHz above the local oscillator’s frequency, the IF signal will include an undesirable image signal interfering with the desired signal. For example, if the desired station is at 600 kHz, then the local oscillator’s frequency is 600 kHz 1 455 kHz 5 1055 kHz. The mixer will output a signal at |1055 kHz – 600 kHz| 5 455 kHz. However, if the antenna receives a radio station at 1510 kHz, and if the BPF tuned to 600 kHz does not suf- ficiently attenuate signals at 1510 kHz, then the output of the mixer will have another signal at |1055 kHz – 1510 kHz| 5 455 kHz. Thus, the RF band-pass filter should be tuned to the desired station and remove substantially all undesirable RF signals that would be inadvertently mixed down to the intermediate frequency. Fortunately, in most AM radios, there is a sufficiently high Q (e.g., narrow bandwidth) tuned circuit that passes the desired RF signal while substantially attenuating the image signal (e.g., 1510 kHz) that is 455 kHz above the local oscillator frequency (e.g., 1055 kHz). The IF filter and amplifier is usually a ceramic filter or an LC (inductor capacitor) band-pass filter with one or more transistors, or a dedicated IC (e.g., TA2003). The output of the IF filter-amplifier is then connected to a demodulator such as an AM (amplitude modulation) detector (e.g., half-wave rectifier circuit). If the system is an FM radio, the detector will employ a frequency modulation detector (e.g., a quadra- ture detector, a ratio detector, or a frequency discriminator). The detector is coupled to an audio amplifier that provides an audio signal to an earphone or loudspeaker.

One disadvantage of the traditional superhet radio is that the system blocks are

fixed to one type of function. For example, if the incoming RF signals also include Morse Code CW (continuous wave) and single sideband signal (SSB) signals, then the detector system will be complex with multiple demodulation circuits (e.g., prod- uct detector with CW filter, product detector with SSB filter, etc.). Although these different types of detectors can be made in hardware, they can be mimicked via a software program in a computer, once signals are digitized through the soundcard. See Figure 12-2.

Some Ham Radio Circuits Related to SDR        269

FIGURE 12-2  A block diagram of an SDR front-end system with low-pass filters for the IF.

As we can see in Figure 12-2, there are some similarities with the standard super- het radio. There’s still an RF band-pass filter, but now we have twice as many RF mixers, twice as many signals for the local oscillator, and twice as many IF amplifiers and filters. Fortunately, these types of mixers and IF filters (low-pass filters, LPFs) can be implemented with op amps and simple resistor capacitor filters. The reason for having twice as many parts after the RF band-pass filter is so the computer can take the two signals, I OUT and Q OUT, and process them such that any image sig- nal will be cancelled out via the software program installed in the computer (e.g., Winrad). Basically, the local oscillator has an I phase signal (defined as 0 degrees phase reference) and a Q phase signal (90 degrees phase shift from the I phase sig- nal). The outputs of each of these I and Q oscillator signals are fed to separate I and Q RF mixers. From the outputs of the I and Q mixer are frequency translated signals of two low-frequency IF signals that are 90 degrees phase shifted from each other in terms of their carrier signals. Figures 12-3 and 12-4 illustrate this. The two low-pass

FIGURE 12-3 An example RF amplitude modulation signal at a high frequency such as

> 1 MHz or more.

FIGURE 12-4 Examples of two lower intermediate frequency (IF) signals where the carrier signal’s phase of the bottom AM signal is 90 degrees shifted with respect to the top waveform. See center vertical axis as a reference.

filters then provide signals whose intermediate frequency is generally within a 20-kHz, 40-kHz, or 96-kHz bandwidth for a computer soundcard.

Now let’s examine a simple mixer SDR front-end circuit that includes the RF band-pass filter, I and Q analog switch mixers, and low-pass filter amplifiers. We will first start with Figure 12-5, the local oscillator that generates I phase and Q phase carrier signals.

A basic crystal oscillator via IC1E utilizes a logic inverter gate configured as an inverting gain amplifier via negative feedback resistor R1 to establish a DC operating point such that the logic inverter gate IC1E is self-biased in an amplifying region. The value of R1 is chosen to be high resistance, with R1 ≥ 1MΩ so that it will not load down the signal from the crystal Y1 at C5. If R1’s resistance is too small (e.g.,

< 1000Ω), the circuit may not oscillate due to excess attenuation at pin 11. The oscil- lator circuit requires a phase lag of about 60 degrees or more from RC filter R8 and C4 so that the crystal Y1 and capacitor C5 form another phase lag circuit of at least 90 degrees (e.g., 90 degrees to 135 degrees). To ensure oscillation, the phase lag from R8 and C4 combined with the phase lag from Y1 and C5 has to be in the order of 180 degrees since the inverter gate “automatically” gives another 180 degrees of shift. With a total phase shift of 360 degrees via the phase shift networks (R8, C4, Y1, C5), an oscillation occurs at IC1E pin 10. The oscillator signal is coupled to a buffer inverter IC1A that drives a binary divider, IC3A, which provides a divide by 2 signal at pin 3 and a divide by 4 signal at pin 4 of the 74HC393 IC. A D flip flop is clocked by the divide by 2 signal (e.g., 7.16MHz) at pin 3 IC3A, which provides one-bit sample-and-hold signals of the divide by 4 signal at pin 2 of the 74HC74, which in

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FIGURE 12-5 A crystal oscillator circuit with flip flops to generate I and Q carrier signals, where Vosc_I is defined as the 0 degrees signal and Vosc_Q is 90 degrees shifted in reference to the Vosc_I signal.

turn provides a quarter-cycle or 90 degrees delayed signal of the original divide by 4 signal. See Figure12-6.

FIGURE 12-6 Top trace is Vosc_I (0 degrees reference) square-wave signal and the bottom trace is Vout_Q (90 degrees), a delayed square-wave signal.

The two oscillator signals, Vosc_I and Vosc_Q, will be used for the I and Q mix- ers as shown in Figure 12-7 where IC5A and IC5B are square-wave RF mixers.

Vin is from a 50Ω antenna tuned to ~ 3.60 MHz for the 80-meter ham radio band. RF matching network Cvar1, L1, and R13 form an RF matching network for Rin 5 50Ω. This RF matching network is a high Q high-pass filter that includes some band-pass filtering around the resonant frequency of about 3.60 MHz. Generally, we want to step up the RF input voltage via Cvar1 and L1 many times. This step-up factor is Q and generally we want Q ≥ 3 so that the subsequent calcula- tion will be accurate for practical purposes (e.g., within 10 percent).

The RF voltage across L1 is basically Q 3 Vin. So, this RF matching network provides an equivalent step-up transformer effect. For example: Q 5 XL1/50Ω, for

3.60 MHz 5 fres and L1 ~ 22 uH, XL1 5 2πfres(L1) so Q 5 2π(3.60MHz)(22 uH)/50Ω or Q ~ 10. We can find Cvar1 from fres 5 3.60 MHz 5 1/[2π=(L1)(Cvar1)], which leads to Cvar1 5 1/[L1(3.60 MHz 3 2π)2] Cvar1 5 1/[22 uH (3.58MHz 3 2π)2] Cvar1 5 [1/(1.112)10] farad or Cvar1 5 90 pf. Cvar1 can be a polyvaricon variable capacitor used in AM radios (e.g., two sections, 0–60 pf and 0–140 pf).

To ensure that Rin 5 50Ω, R13 is required to “terminate” at L1 so at the resonant frequency of 3.60 MHz, Rin is resistive and at 50Ω. Having Rin 5 50 Ω is important so that the antenna’s coaxial cable is properly loaded/terminated to ensure low SWR (standing wave ratio), which means maximum power from the antenna is trans- ferred to the front in circuit at Vin via Cvar1. To calculate R13, you can use the fol- lowing formulas for Q ≥ 3.

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FIGURE 12-7 An SDR front-end circuit with input band-pass filter (Cvar1 and L1) amplifier Q1, I and Q mixers (IC5A and IC5B) and low- pass filter amplifiers (R4, C12, IC6A; R5, C13, IC6B).

Note that Cvar1 is adjusted for maximum amplitude at transistor Q1’s emitter since L1’s inductance is accurate to about 10 percent to 20 percent.

R13 5 Q 3 XL1 5 10 3 2πfres(L1) 5 10 3 2π(3.60MHz)(22 uH), or R13 ~ 5000Ω.

When you know Q, another faster way of calculating R13 is to use the following:


Q 5 =R13, where Rs 5 input RF signal’s source resistance, which is generally 50Ω.

For example, Rs 50Ω.                                                                                                     

This leads to: R13 5 Q2 3 Rs. Q 5 =(R13/Rs) 5 =(5000Ω/50Ω) 5 =100 5 10

Since Q 5 10:

R13 5 102 3 50Ω 5 100 3 50Ω or R13 5 5000Ω

If higher Q is desired, we can raise the L1’s inductance such as L1 5 47 uH, then Q 5 23.7. This will lead to having Cvar1 5 42 pf and R13 5 (23.7)2 3 50Ω or R13 5 28KΩ. (Note: You can use 27KΩ, which is close enough.)

Note that Q1 does not have infinite input resistance to its base terminal. This input resistance at Q1’s base is roughly β 3 (R11 || ZL2), which will be in parallel with R13. For an approximation given R11 5 1KΩ and |ZL2| 5 2π(3.6 MHz)L2 5 2π(3.6 MHz)330 uH 5 7.46KΩ >> 1KΩ. So β 3 (R11 || ZL2) ~ β(R11) 5 β(1KΩ).

Since β ranges between 50 to 100, the input resistance is about 50KΩ to 100KΩ,

which is in parallel with R13. Given R13 in these two examples can be 5KΩ or 28KΩ, a good approximation is just to increase R13 by 20 percent from the calculated value. Thus 5KΩ  6KΩ (6.2KΩ for a standard resistor value), and 28KΩ  33.6KΩ (33KΩ or 36KΩ). Thus, slightly increasing R13’s calculated resistance compensates for Q1’s input load resistance.

NOTE: In many instances, the antenna may just be a long wire connected to Vin or Cvar1 directly. Here, the source resistance of the long wire antenna may not be 50Ω, and it is possible that R13 can be removed to maximize gain at the resonant frequency.

Emitter follower Q1 acts as a unity gain amplifier with high input resistance at the base of Q1. With L1 providing a 0-volt Q1 DC base bias voltage, Q1’s emitter DC voltage is –0.7 volt since VBEQ1 5 10.7 volt. If you are not sure how the DC voltages for this Q1 emitter follower are calculated, you can build it with L1, Q1, R12, R11 with

± 5-volt supplies and measure the emitter voltage with a voltmeter with the negative test lead connected to ground.

The DC collector current is set by R11, ICQ1 ~ (–0.7 v – –5 v)/R11 ~ 4.3 v/1000Ω or ICQ1 5 4.3 mA. However, the DC collector current can be lowered to as low as ~ 2 mA or R11  2200Ω, because the input signal will be very small (< 100 mV) and does not require large signal swings at Q1’s emitter. The output of Q1 is fed through a coupling capacitor C9 into a large value inductor that establishes 0 volt DC at the

“Y” inputs of the 74HC4053 analog switches at pin 13 and pin 1 of IC5A and IC5B. Note that series resistor R10 5 47Ω provides some isolation from the “kickback” spikes or glitches that will emanate to the input terminals at pins 13 and 1 due to switching characteristics of the 74HC4053. In general, when using an analog switch as an RF mixer, it is always good to add a small value series resistor to the input so that the amplifier driving it (e.g., Q1 in this case) does not exhibit instability due to the dynamic loading caused by the 74HC4053 input terminals’ glitch signals. Note that there is an alternative to lowering the amount of glitch kickback by choosing a lower capacitance analog switch such as the SD5000 series FET (Note that the SD5000 or SD5200 quad switch has a different pin out than the 74HC4053).

Now let’s take a look at the mixers, IC5A and IC5B, that gate through the RF input signal via Q1’s emitter at a rate of the local oscillator’s frequency. The output of each of the mixers at pins 14 and 15 will contain signals related to RF input signal, an IF signal related to the difference of the frequency of the RF signal and the frequency of the local oscillator, plus other high-frequency signals beyond the frequency of the RF input signal. See Figures 12-8 and 12-9.

FIGURE 12- 8 Top trace is the RF input AM signal at 3.6 MHz carrier at 1 kHz modulation; bottom trace is a mixer output at IC5A pin 14. Note the two amplitudes are the same.

FIGURE 12-9 Top trace is the same RF input signal; bottom trace is the low-pass filtered version at R4/C12. Note that the filtered version is about one-third the amplitude of the RF input signal.

With the local oscillator providing 3.58 MHz and with a 3.60 MHz RF input signal, the IF (intermediate frequency) will be |3.60 MHz – 3.58 MHz| 5 0.02 MHz or the IF 5 20 kHz.

As we can see in Figure 12-9, when all the high-frequency components are fil- tered out of the amplitude of the IF signal, we see a signal that still retains the type of amplitude modulation envelope of the RF input signal. However, the amplitude is reduced to about one-third. That is the conversion gain 5 amplitude of the IF signal after filtering (e.g., at R4/C12) divided by the amplitude of the RF input signal at the input of the mixer (e.g., pin 13 of IC5A).

Theoretically, for the IC5A/B mixer using Fourier Series (which the reader does not need to know), the conversion gain (1/π) 31.847%, which is close to 33 percent or one-third.

Amplifiers U6A and U6B are non-inverting gain amplifiers each with a gain of (1 1 47K/1K) or a gain of 48. The total conversion gain from the input of the mixer at L2 or pins 13 and 1 of IC5A/B to the output of either U6A or U6B is then (1/π) 48 15.2.

NOTE: The op amp chosen in this example is the OPA2134, which has an 8 MHz gain bandwidth product. So, for a gain of 48, the resulting bandwidth will be 8 MHz/48 ~ 166 kHz, which is sufficient for an audio sound card that accepts a 96 kHz analog input bandwidth using a 192-kHz sampling rate.

Some Troubleshooting Tips Concerning Figure 12-5 and Figure 12-7

1.            Check wiring on the pin outs for all the ICs. If an IC is loaded in reverse, you may burn it out because the power and ground pins are at pins 14 and 7 or at pins 16 and 8. Reversing the IC will supply positive voltage to the ground pin and ground the 1 power pin, which will damage the chip. If this happens, shut off the power and carefully remove the chip and put in a new one with the correct orientation. It generally helps to use IC sockets instead of soldering the chip directly into the circuit board.

2.            Confirm that the power supplies 15 volts and –5 volts are at the appropriate pins for IC5, IC6, the collector of Q1, and emitter resistor R11. And confirm 15 volts on the power pins of IC1, IC3, and IC4. Also, with an ohm meter, confirm the ground terminal pins for IC1, IC3, IC4, and IC5. Also, make sure that the power supply decoupling capacitors C10, C14, C15, and C16 are each located near (within half an inch of) the power pins to the ICs. Confirm that these decoupling caps are making continuity to the ICs’ power pins (e.g., pin 14 for a 14-pin chip and pin 16 for a 16-pin chip), and that the other sides of the decoupling caps are connected to ground with short leads. The ground connection of a decoupling capacitor can be made via a ground plane on a printed circuit board or to pin 7 of a 14-pin chip or for a 16-pin chip to pin 8.

3.            For the oscillator board in Figure 12-5, confirm that the crystal oscillator is working with an oscilloscope. Make sure that the oscilloscope is set to a bandwidth of at least 50 MHz bandwidth, otherwise a 14.318-MHz square wave will look too sinusoidal. Confirm a 14.318-MHz sine wave of at least 1 volt peak to peak at pin 11 of IC1E, and an approximate square wave at IC1E pin 10 and IC1A pin 2 of about 5 volts peak to peak also at 14.318MHz. If you see noise at IC1E pin 10, check if R1 (2.2MΩ) is the correct value and that it is connected as shown in Figure 12-5; or try another crystal. Also, check R8 (2200Ω) for the correct resistance, and confirm that C4 and C5 are 22 pf.

4.            Check for a 5-volt peak to peak square wave at IC3A pin 3 that should be one- half the frequency (7.16 MHz) from IC1A pin 2, and confirm that pin 4 of IC3A has a one-quarter frequency (3.58 MHz) square wave at about 5 volts peak to peak. If you see no output, confirm that pin 2 of IC3A is grounded.

5.            With an oscilloscope, check for 90 degrees difference in waveform in Vosc_I at pin 2 of IC4A and Vosc_Q at pin 5 of IC4A. See Figure 12-5. If signal Vosc_Q is not a 90-degree phase-shifted square-wave signal relative to Vosc_I, then make sure pin 1, pin 4, and pin 14 of IC4A have 15 volts on all of them, and that pin 7 of IC4A is connected to ground.

6.            At this point if you see very small amplitude waveforms, check to see if any of the output pins such as IC1E pin 10 and IC1A pin 2, IC3A pins 3 and 4, or IC4A pin 5 are shorted to ground or shorted to 15 volts.

7.            For the SDR front-end board in Figure 12-7, we can confirm that the oscillator signals, Vosc_I and Vosc_Q, are at IC5A pin 11 and at IC5A pin 10, respectively.

8.            Connect a 50Ω source resistance generator at 3.60 MHz and 200 mV peak to peak into the Vin terminal at Cvar1. With an oscilloscope probe at the emitter of Q1, confirm at least 1.5 volts peak to peak when Cvar1 is adjusted for maximum amplitude. The gain should be close to 10 ± 20 percent from Vin to the emitter of Q1. You may need to set your scope to “Bandwidth Limit” and AC coupling. Bandwidth Limit reduces the analog signal bandwidth to 20 MHz in most oscilloscopes but allows for viewing cleaner waveforms. Since the frequency is 3.6 MHz, the 20-MHz bandwidth limitation does not interfere in observing waveforms accurately for the SDR front-end circuit. Usually, you can obtain an accurate frequency (< 0.05 percent error) digitally synthesized waveform generator for less than $100.

9.            Confirm sine waveforms at pins 13 and 1 of IC5 that are at least 1.4 volts peak to peak at 3.60 MHz.

10.            Confirm sine waveforms at pin 3 and pin 5 of U6, which should be about one- third the amplitude of the waveform at pins 13 and 1 of IC5. For example, we should see at least 1.4 volts/3 or at least 0.46 volt peak to peak at about 22 kHz at pins 3 and 5 of U6.

11.            The gain of U6 is about 48, so we will get clipping at the outputs pin 1 and pin 7 of U6. Thus, turn down the RF generator from 200 mV peak to peak to 40 mV peak to peak at 3.60 MHz. With a reduction by fivefold, we should get about 0.092 volt peak to peak at pins 3 and 5 of U6. The output of U6 pins 1 and 7 should now have 48 3 0.092 volt peak to peak or 4.4 volts peak to peak sine wave at about 22 kHz. Confirm that V_I and V_Q have a phase difference of about 90 degrees.

12.            If there are problems with getting waveforms in the “ball park,” confirm the component values of R4, R5, C12, C13, and R8, R9, R6, and R7. For example, here’s a test for the bandwidth of the low-pass filters R4/C12 and R5/C13. You can sweep the RF frequency, fRF, of the input RF signal from 3.59 MHz to 3.69 MHz while probing low-pass filter output at R4/C12 and low-pass filter output at R5/C13 to confirm that the frequency response is “flat” from 10 kHz to about 40 kHz and has a –3 dB amplitude point (70.7 percent

reference) at 100 kHz. Remember that the resulting frequencies from the down-converted IF (intermediate frequency) signal at the low-pass filters is

|fRF – fLO|, where fLO 5 local oscillator frequency, which is 3.58 MHz 5 fLO. Actually, the precise local oscillator frequency is 3.579545 MHz ~ 3.58 MHz, so more accurately, the IF 5 |fRF – 3.579545 MHz|.

A Common Sample-and-Hold RF Mixer Circuit

The switch mode RF mixer shown in Figure 12-7 has the advantage that it does not load the input signal much because of the series resistors in the low-pass filter cir- cuits, R4 and R5 (both 3000Ω), which can be made larger for less loading if needed. For example, if the same bandwidth is required, we can just keep the same RC time constant, which is R4 3 C12 or 3000Ω 3 470 pf 5 1.41 μsec. So, for instance, we can have R4 5 R5  10KΩ and C12 5 C13  140 pf. This way, the loading at the input terminals pins 13 and 1 of IC5A/B will be 10KΩ, worst case. Again, see Figure 12-7. The IC5A/B mixer can be made to switch very fast if other analog switches are used such as the SD5000 quad FET (Field Effect Transistor) or if two FST3253 ICs are used. Note that neither SD5000 nor the FST3253 devices are pin for pin compatible with the 74HC4053 chip.

However, the conversion gain of the IC5A/B in Figure 12-7 is “lossy” because it has about a 32 percent conversion gain (see Figures 12-8 and 12-9). If the RF mixer is a sample-and-hold circuit where the sampling time is relatively small, such as

≤ 25% 3 (1/fLO), then the conversion gain goes up to ≥ 80 percent. See Figure 12-10. In this section we will explore the Tayloe RF Mixer, which is based on sample-and- hold circuits.

FIGURE 12-10 A simple sample-and-hold circuit; note that the sampling pulse < 50 percent duty cycle.

A simple sample-and-hold circuit is modeled by a finite source resistance, Rsource, such as a 50Ω source from Vin. However, Rsource should also include in series the ON resistance of the analog switch. For example, if Vin has a 50Ω source resistance (such as from a 50Ω antenna) and U1B, has an ON resistance of 20Ω, then Rsource 5 50Ω 1 20Ω 5 70Ω. As we will find out soon Rsource plays an important part in terms of using an analog switch as an RF mixer. The basic operation of the sample-and-hold circuit is when V_LO’s signal is logic high, the analog switch closes and connects for a short duration signal from Vin through Rsource to charge up

capacitor C1. When V_LO goes to logic low, the analog switch becomes an open circuit, which disconnects Vin and Rsource from C1. Then capacitor C1 retains the charge and provides a constant voltage during the logic low state from V_LO. See Figure 12-11.

FIGURE 12-11 An example output signal from a sample-and-hold circuit on the bottom trace versus the input signal on the top trace. The stair-step waveform (bottom trace) from the sample-and-hold circuit maintains the same voltage until the next sampling pulse comes along.

See Figure 12-12 for magnified view of the stair step waveform.

FIGURE 12-12 A closer look and the sample-and-hold waveform with input signal at the top trace and the sampled-and-held waveform on the bottom trace.

Note that frequency of the local oscillator or V_LO is close to the frequency of Vin, the RF input signal. The output at Vout is then an IF signal that has a low frequency due to |fRF – fLO| 5 IF and fRF almost equal to fLO. For example, we want

|fRF – fLO| < 100 kHz so that the IF signal at Vout can be sent to a sound card in a

computer to run an SDR program. If we are trying to down-convert an 80-meter

ham radio RF input signal frequency fRF 5 3.800 MHz, the local oscillator signal, V_LO, can have a frequency of fLO 5 3.810 MHz such that: |fRF – fLO| 5 IF 5

|3.800 MHz – 3.810 MHz| 5 |–0.01 MHz| 5 0.01 MHz or the IF 5 10 kHz.

In general, the sampling pulse (high logic state duration) should be << 50 percent duty cycle. That is, if you have a 1-MHz sampling frequency, the period is 1 μsecond (μsecond 5 μsec). The duty cycle is the “ON” time divided by the period. In this example, it will be ON time/1 μsec. Normally, you would like to have the ON time as short a duration as possible. In sampling theory the ON time duration  0 μsec, but this is not practical because with Rsource in series with sampling capacitor C1, the capacitor C1 in Figure 12-10 cannot charge up instantaneously. So, we need to first define what the duty cycle will be. Then second, determine what we need for a good C1 capacitance value to use given that Rsource is known. To get a feel of the effects of too much capacitance in C1 see Figure 12-13, which shows examples of pulse responses with “too much capacitance” on the longer RC time constant waveform compared to the shorter RC time constant waveform that provides the final voltage with a small time interval.

FIGURE 12 -13 With different RC time constants we see that it takes a longer sampling pulse (t) to arrive at the final voltage V0. And ~ V0 is achieved at t1 with a shorter RC time constant when compared to ~V0/2 on the longer RC time constant waveform.

The actual equation for the voltage across the capacitor as the analog switch is closed for the duration a positive going pulse from 0 volts to V0 volts is:

Vout 5 V0(1 – e–t/RC)

where V0 5 peak to peak amplitude of the pulse and where the pulse starts at 0 volts, t 5 time duration of the sampling pulse (e.g., time duration of the V_LO in the logic high state), and where R 5 Rsource and C 5 C1 as shown in the circuit of Figure 12-10, and where 5 2.71828.

Ideally, we want Vout to have the same value as V0 instantly. But this will not happen due to the finite resistance, Rsource. So, here’s what the charge-up voltage at Vout looks like in terms of RC time constants where “t” is the ON time duration:

Vout 5 63% of V0 when t 5 RC Vout 5 86% of V0 when t 5 2RC Vout 5 95% of V0 when t 5 3RC

For example, suppose we want to build a sample-and-hold mixer for the 80-meter band around 3.600 kHz, and we want to use a 25 percent duty cycle base on a local oscillator frequency at 3.600 MHz, whose period for one cycle is 1/(3.6 MHz) 5

0.277 μsec or 277 nsec (nano seconds). For a 25 percent duty cycle, the logic high “ON” duration will be 25% 3 (277 nsec) 5 69 nsec. If we want to have Vout come up to 95 percent of input value, we have to set the RC time constant to one-third of the “ON” duration or (1/3) (69 nsec) or t’ 5 RC 5 23 nsec. This means if Rsource 5 50Ω, then C1 5 t’/Rsource 5 (23 nsec)/50Ω or C1 5 463 pf ~ 470 pf. We label this as C180meters 5 463 pf. Note: For 86 percent of input value set RC to (1/2) (69 nsec).

But what if we want to receive other ham radio bands up to 20 meters? Then we

simply scale C1 accordingly to a smaller value in proportion to C1 for 80 meters.

That is, C120meters 5 (20 meters/80 meters) C180meters.

Or for 20 meters, C1 5 (1/4) 463 pf or C1 5 115 pf ~ 120 pf or 100 pf.

But what this really means also is that you can use the 20 meters C1 value for any lower frequency. For example, if C1 5 100 pf in Figure 12-10 with Rsource 5 50Ω, this means for the lower frequencies (e.g., 40 meters, 80 meters, and 160 meters) the pulse duration will be longer and Vout will be greater than 95 percent (but always less than 100 percent).

Now let’s look at the consequence of having too large of a C1 value when we want to measure the IF bandwidth or roll-off. That is what happens when the RF input signal’s frequency starts separating more from the local oscillator frequency. Will the IF signal’s amplitude have the same output voltages at 5 kHz versus 20 kHz or 100 kHz? The answer is no. The larger the C1 capacitance, the poorer the frequency response will be on the higher frequency end of the IF signal. See Table 12-1, where Rsource is 50Ω that represents a “best” case scenario. There are times when the source resistance is much higher, such as when connecting a single “untuned” long

wire antenna directly into the RF mixer without an amplifier. Such resistances can be much higher, such as 300Ω or more. Table 12-1 will “stop” at the –3 dB frequency for each C1 value from Figure 12-10. The lowest frequency, 100 Hz, is defined as 100 percent with the 0.01 uf 5 C1.

The test setup has the local oscillator signal, V_LO, at 3.6000 MHz with a 25 percent duty cycle (69.4 nsec high logic level for a one-cycle period of 277.7 nsec). Vin is a signal at 3.6000 MHz plus the frequencies shown in Table 12-1.

For example, to provide a 1000 Hz IF signal at Vout, the input RF signal is 3.6010 MHz, such that 3.6010 MHz – 3.6000 MHz 5 0.001 MHz 5 1 kHz.

TABLE 12-1 Comparing Frequency Response of the IF Signal at Vout of Figure 12-10, a Single Sample-and-Hold Circuit with Different Hold Capacitor Values for C1 and with Rsource = 50Ω

 Vout Freq.% Vout @ C1 = 0.01μf%Vout @ C1 = 0.22μf–3 dB (70.7%of Max)
100 Hz100%96% 
1000 Hz99%95% 
5000 Hz97%66%4900 Hz C1 = 0.22uf
10 kHz96%41% 
20 kHz94%24% 
30 kHz90%17% 
90 kHz71%— (too small to measure)90 kHz C1 = 0.01uf

As we can see, there is a penalty in having a large capacitance hold capacitor C1. With the 0.22 μf for C1, the usable bandwidth is about 5 kHz, which is about one- fourth the bandwidth most minimum sampling rate sound cards can do at 44.1 kHz, which provides an analog signal bandwidth of 20 kHz. Reducing the capacitance from 0.22 μf to 0.01 μf is much better and a 90-kHz bandwidth is achieved when Rsource 5 50Ω. However, if Rsource > 50Ω, such as 300Ω, then we are back to a reduced bandwidth situation. Also, if we want to listen in on higher frequency signals such as in the 40- or 20-meter ham radio bands, then the high logic level “ON” time reduces accordingly to about 35 nsecs or 17.5 nsecs. Thus, the 0.01 μf value for C1 will start providing some more bandwidth loss due to the shorter sampling intervals.

For RF sample-and-hold mixer circuits, the hold capacitor should be in the range of 47 pf to 1000 pf; this will allow for providing a good IF bandwidth up to 90 kHz or more, which can be taken advantage of with a high-quality sound card with 192 kHz sampling frequency. With higher IF bandwidth, more of the radio band can be tuned with a fixed frequency local oscillator.

In general, when using a software defined radio (SDR) program such as Winrad (or others), the span of tuning across the radio band is related to twice the IF band- width. Recall that in Figure 12-2, there are two mixers, one I and one Q. Both mixers

will provide the same IF bandwidth, and if the hold capacitor has too much capaci- tance, then the tuning range within the SDR software program will be restricted.

For example, with C1 5 0.22 μf for each of the sample-and-hold mixers (I and Q), the IF bandwidth is ~ 5 kHz. This means only 2 3 5 kHz or 10 kHz of tuning across the band is possible. If C1 5 0.01 μf for a 90 kHz IF bandwidth, then we get 2 3 90 kHz or 180 kHz of tuning range. This then is eighteen times better than the 5 kHz IF bandwidth.

A Preferred Implementation with Sample-and-Hold Circuits

As have already seen in Table 12-1, using a large-value hold capacitor dramatically reduces the IF bandwidth that hinders tuning range. With all sample-and-hold cir- cuits, an important principle to keep in mind is the hold capacitor such as C1 in Figure 12-10 needs to be connected to a high impedance input amplifier. That is, any (e.g., low to medium) resistive loading (e.g., ≤ 10kΩ) across the hold capacitor (e.g., C1) will discharge the capacitor’s voltage before the next sample pulse and result in a loss of output signal voltage.

In some cases, the type of amplifier used to amplify the capacitance voltage at C1 of Figure 12-10 should be FET or amplifiers with extremely low input bias currents (e.g., ≤ 0.200 nA for JFET op amps). In general, almost any FET op amp will work, such as the LF353, TL082, TLC272, etc., providing you keep track of the gain and IF bandwidth.

For example, if you want a 100-kHz closed loop bandwidth from the op amp non-inverting gain amplifier and the gain bandwidth product is 3 MHz, then the maximum gain you can have is about 130 since 130 3 100 kHz 5 3 MHz.

Figure 12-14 shows a preferred way of providing amplification to a sample-and- hold circuit, and Figures 12-15, 12-16, and 12-17 illustrate ways of causing problems to the sample-and-hold circuit that should not be used unless corrected or modified. The voltage across C1, Vout, preferably should be amplified or buffered with a high input resistance amplifier (e.g., > 1 MΩ). If an op amp is used, this means that only non-inverting gain amplifier configurations will work. At first look, Figure 12-14 may seem “funny” or incorrect because there is no bias resistor to set the DC bias for the (1) input terminal of the op amp. The reason we can get away from hav- ing a bias resistor at the (1) input is that the analog switch, U1B, will transfer a

sample of the input voltage to C1, which in turn biases the (1) input terminal.

In Figure 12-14, note that the input signal Vin must have a DC path to a DC volt- age source or to ground. The gain of the amplifier is Vout2/Vin 5 [1 1 (R1/R2)]. Generally, the gain is in the order of 10 to 100 in most cases. With higher bandwidths such as having a gain of 100 with an IF bandwidth of 96 kHz, you can choose an op amp (e.g., TLE082) with a gain bandwidth product of ≥ 10 MHz since 100 3 96 kHz equals 9.6 MHz. See the end of this chapter for more on gain bandwidth product.

FIGURE 12-14 A preferred way of buffering or amplifying voltage across a hold capacitor to keep the hold capacitor’s voltage at C1 steady until the next sampling pulse arrives via V_LO.

Alternatively, you can split the gain of 100 by cascading two amplifiers by having each non-inverting gain op amp stage provide a gain of 10. This way, you can get away with an op amp with a gain bandwidth product of about 1 MHz. However, it’s best to be on the safe side and use commonly available op amps such as the TLC272, TL082, or LF353 that have gain bandwidth products of > 2MHz. The penalty for this is adding more op amps, but usually these op amps are less costly than the high- speed versions.

AC bypass capacitor, C2, can be omitted with R1 grounded on the R1/C2 side. However, C2 may be used to avoid amplifying DC voltages. Recall the input signal into the computer for the SDR program typically will be AC coupled into the sound card. If Vin has a DC bias voltage, then the op amp with C2 will just pass the same DC offset voltage to Vout2 without amplifying the DC offset voltage that can cause the op amp to saturate or clip at its output Vout2.

In Figure 12-15, there is no need to include R3 in the circuit, especially if this resistor discharges C1 too quickly. Setting a value such as < 50KΩ (e.g., R3 5 10KΩ or R3 5 1KΩ) will cause the sample-and-hold circuit to have a reduced output volt- age at Vout2. If resistor R3 were a higher resistance value in the range of 1MΩ to 10MΩ, then there would generally be a minimal signal amplitude reduction at Vout2, and this change into much higher resistance values will fix the problem. However, you can just remove R3.

We now turn our attention to another error with sample-and-hold circuits that are connected to inverting gain amplifiers. One of the most “major” mistakes is to connect the hold capacitor (e.g., C1) to an inverting gain operational amplifier cir- cuit that has a low input resistance. This low input resistance will then discharge the hold capacitor that results in an attenuated output voltage. See Figure 12-16.

FIGURE 12-15 A non-inverting gain amplifier with a 10kΩ resistor R3 across hold capacitor C1. R3 prevents C1 from doing its job of holding a steady voltage until the next local oscillator sampling pulse arrives. To correct the situation, R3 may

be removed.

FIGURE 12-16 A sample-and-hold circuit that will definitely not work as planned because the hold capacitor is connected to an inverting gain amplifier that has a low input resistance via R1.

In Figure 12-16, the load resistance as seen by the hold capacitor, C1, is Rin, which is an equivalent resistor that is in parallel to C1. The reason for this is because the (–) or inverting input terminal of the op amp U2A is a virtual ground due to its

(1) input terminal (e.g., pin 3) being connected to ground. Typically, R1 will be ≤ 1KΩ (e.g., R1 has a range of 10Ω to 220Ω in some circuits found on the web or Internet). This very low resistance will then cause the hold capacitor C1 to discharge almost fully or substantially before the next sampling pulse arrives in the V_LO sig- nal. Because the IF signal is attenuated at C1 (Vout) due to Rin’s low input resistance

of R1, the gain has to be pumped back up. So typically, R1 ~ 10Ω and R2 ~ 5KΩ for a gain of 500. Because of the high gain a required IF bandwidth of up to 96 kHz, the op amp U2A has to be a high-speed op amp with a gain bandwidth product of ≥ 50 MHz. So, a 3 MHz TL082 as shown inFigure 12-13 will not work very well and result in a closed loop bandwidth of 3 MHz/500 5 6 kHz that yields a 6 kHz IF bandwidth.

NOTE: The sampling pulses as shown in Figure 12-16 show about a 25 percent duty cycle. For only Figures 12-16 and 12-17 more conversion gain may be had if V_LO were 50 percent square waves. At this point the hold capacitor, C1, does almost nothing since the low impedance of Rin diverts most of the IF signal current away from C1.

Our next circuit in Figure 12-17 shows another added problem when the input resistor R1  0Ω, which can cause an op amp to oscillate due to hold capacitor C1 adding extra phase shift.

FIGURE 12-17 A sample-and-hold circuit that will not work as planned because the hold capacitor C1 is connected directly to the inverting input of the op amp that can cause parasitic oscillations from the op amp.

By having a hold capacitor, C1, connected directly to the inverting input of the op amp with feedback resistor, R2, an oscillation can occur. The reason is that R2 and C1 form a low-pass filter that causes a phase lag from Vout3 to pin 2, the (–) input terminal. This extra phase shift along with the phase shift of the op amp can then cause a net positive feedback situation, which in turn produces oscillation. Only by “luck” of picking the op amp and having an R2 3 C1 time constant that does not have sufficient phase shift is there no oscillation. If other capacitance values for C1 are chosen or if a different op amp is used, then oscillation at Vout3 can occur. In summary, Figure 12-17 represents a circuit “waiting for a problem.”

And of course, the input resistance, Rin, is very low. In general, the |Rin| ~ R2/[a(f)], where a(f) is the open loop gain of the op amp. So even if R2  5KΩ, we know that Rin < 5KΩ because a(f) is generally much greater than 1 (e.g., a 50 MHz

op amp will still have an open loop gain of 10 at 5MHz, and an open loop gain of 100 at 500 kHz, and an open loop gain of 1000 at 50kHz). For example, if a 50 MHz op amp U2A is used such as an LM4562, and R2 5 10kΩ, and the IF or mix down fre- quency is 50 kHz:

Rin 5 R2/(open loop gain at 50 kHz) 5 10kΩ/1000 or Rin ~ 10Ω

So again in Figure 12-14, the hold capacitor, C1, cannot hold a constant voltage from one sample pulse to another (e.g., see Figure 12-14 where the hold capacitor loading into infinite or high resistance (e.g. ≥ 1 MΩ) results in a constant voltage before the next sample pulse and a stair step waveform (flat steps) is produced).

A Cool Four-Phase Commutating Mixer

Figure 12-7 showed a simple I and Q mixer for software defined radios (SDRs). We will now look into the terrific Tayloe four-phase RF mixer that can be used for gen- erating I and Q signals for SDRs. First let’s look at a basic configuration. See Figure 12-18.

FIGURE 12-18  A Tayloe RF mixer that provides I and Q signals.

In some of the most basic configurations, input capacitor C1 (e.g., C1 5 1 μf) serves as an AC short circuit and blocks DC voltages from Vin. However, it is better for having L1 and C1 forming a basic high-pass filter that removes low-frequency interference signals. This high-pass filter may be a broadband filter such as rejecting signals below several hundred kHz, while passing RF signals above 500 kHz. Generally, as with any sampling or switch mode filter, an RF band-pass filter is desir- able before the RF mixer to ensure that any out-of-band signal does not down-con- vert to an interfering IF signal. For example, we can also select values for L1 and C1 as a high Q (e.g., Q ≥3) high-pass filter by having [2πfres (L1)]/50Ω ≥3, where

fres 5 1/[(2π=(L1)(C1)]

and where Vin (e.g., an antenna) has a source resistance of 50Ω.

This mixer operates by sampling and holding the (four) signals between four equally spaced intervals that are one-quarter cycle apart. So, there is a four-position sequence of 0 degrees, 90 degrees, 180 degrees, 270 degrees, and then the sequence repeats over time. For example, in a 160-meter receiver at 1.8000 MHz, the switch control signal, Vsw-control, provides a sampling pulse for each phase at a 1.8000 MHz rate. With an equivalent circuit using four separate switches, Figure 12-19 shows the “ON” time when the multiplexor, MUX gates Vin’ to each of the capaci- tors, C_0, C_90, C_180, and C_270. Figure 12-19 shows that each capacitor is charged via Vin’ one at a time and just as one pulse ends, another starts via Vsw_0, Vsw_90, Vsw_180, and Vsw_270. The period of each pulse waveform will still be [1/ (1.8000 MHz)] 5 0.555 µsec in this example for a frequency of 1.8000 MHz.

FIGURE 12-19 Timing pulses for four sampling switches. This circuit mimics MUX1 in Figure 12-18.

Now getting back to Figure 12-18, the voltage across each capacitor is connected to an input terminal of a differential amplifier (e.g., DA1 or DA2), which can be implemented with instrumentation amplifiers (e.g., INA163). Ideally, the input resistance to each input (e.g., Rin_0 to Rin_270) is infinite or in the order of ≥ 1MΩ so that the hold capacitors C_0, C_90, C_180, and C_270 do not discharge, causing a drop in voltage before the next sampling pulse comes along.

The output of DA1, Vout_I 5 k(V_0 – V_180), and the output of DA2, Vout2 5 k(V_90 – V_270), where k is the gain of the differential amplifier (or instru- mentation amplifier).

Now let’s take a look at a circuit that can be improved as shown in Figure 12-20. Some circuits posted on the web (world wide web) are similar to the one in Figure 12-20. In some cases, R7 5 R8 have a range from 0Ω to 220Ω. The circuit will still provide I and Q signals at Vout_I and Vout_Q. However, the amplitude signal levels at V_180 and V_270 will be very small due to the hold capacitors, C4 and C6, being loaded to a low-resistance load via R7 and R8, whereas the two other hold capacitors, C5 and C7, are not loaded at all since these two capacitors are connected

to the high resistance non-inverting inputs of U4 and U5.

The frequency bandwidth at V_0 and V_90 will be low, in the order of less than 10 kHz. This is because the hold capacitors (C5 5 C7 5 0.27 µf) have large capaci- tances and form a low-pass filter effect. As for the voltages at V_180 and V_270, we will see that they will produce “unexpected” signals and they will not be the expected phase inversion signals of V_0 and V_90.

The V_0 and V_90 signals are on the non-inverting inputs (pin 3) of both op amps U3 and U4. Since the inverting input terminal voltages must match the non- inverting input voltages, the signal voltages at V_180 and V_270 are “overridden” by V_0 and V_90. And the voltage at the inverting input terminal is like a low-imped- ance voltage source since the inverting input signal matches the signal at the non- inverting input terminal. Again, see Figure 12-20.

For example, you can think of a voltage follower circuit via the inverting input terminal pin 2 of U3 is driving C4 via low-resistance resistor R7. So, if you look at the signal voltage V_0 at C5, you will see almost the same signal voltage V_180 at C4. Likewise, the V_90 voltage you see at C7 will be approximately the same as the volt- age at V_270 at C6. Remember in an op amp circuit with negative feedback, it is the voltage at the non-inverting input terminal (e.g., pin 3) that determines what the voltage will be at the inverting input terminal (e.g., pin 2). That is, the voltages at pin 3 (positive input) and pin 2 (negative input) will be the same.

For example, if we look ahead at Figure 12-23, the signal voltages at C4 and C6 are “washed out” and you will not see the expected inverted phase signals there. In essence, the inverted phase signals V_180 and V_270 are “thrown away” in this cir- cuit. But we can fix this by using an instrumentation op amp (see Figure 12-25) or by using voltage followers (see Figure 12-27) that will amplify each hold capacitor’s (C4, C5, C6, and C7) voltage with high-resistance input terminals.

In terms of troubleshooting this circuit (Figure 12-20), the first place to start is at V_CLK, which is 3 volts to 5 volts peak to peak pulsed waveform such as a square wave signal. At pin3 of U2A and pin 11 of U2B, you should confirm the clock signal. To determine the frequency, you need to know which radio frequencies you want to receive via Vin. V_CLK will have a frequency 4x the incoming RF frequency. For example, if you want to listen in on the 80-meter band around 3.6000 MHz,

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FIGURE 12-20 An example four-phase RF mixer that can be improved. Capacitors C4 and C6 can cause op amps U3 and U4 to oscillate if R7 = R8 = 0Ω.

V_CLK will have a frequency of 4 3 3.6000 MHz or 14.400 MHz. A V_CLK genera- tor may come from a frequency synthesizer chip with crystal oscillator or from a crystal oscillator. If frequency generators are using an LC (inductor capacitor) or RC (resistor capacitor) oscillators, the frequency stability is most likely inadequate. Fortunately, today you can purchase an inexpensive crystal-controlled, two-channel variable frequency signal generator for testing these types of circuits. Again, see Figure 12-20.

For digital counter/divider chips (U2A and U2B) the first thing to check for is that there is 15 volts at the power pin 14, and pin 7 is tied to ground. Then confirm that the preset and reset pins are tied to logic high (e.g., 15 volts). We should see 15 volts at pins 4 and 10 for the preset pins and pins 1 and 13 for the reset pints. Second, confirm the (clock) CLK pins 3 and 11 U2A and U2B are tied together with a pulsed waveform on these pins, that is, in the 3-volt to 5-volt peak to peak range. Next, confirm with a two-channel oscilloscope with U2A pin 6 connected to Channel 1 and pin 9 of U2B to Channel 2 (of the scope), that you see 5-volt peak to peak waveforms that are 90 degrees phase- shifted from each other. Also, these two waveforms, when starting from both signals at 0 volts, should have a sequence of 0, 2, 3, and 1 where pin 6 of U2A is the least significant bit, and pin 9 is the most sig- nificant bit. Confirm these two waveforms are at U1 pins 14 and 2 (Figure 12-21). Also note the waveforms are not always clean pulses and you may see some small overshoot or slight ringing, which should not be a problem with logic circuits.

In Figure 12-21 the bottom waveform is one-quarter cycle or 90 degrees shifted from the top waveform. Also, note the repeating binary count sequence from the top waveform for the least significant bit and the bottom waveform for the most significant bit. Two-bit binary numbers equal the following in decimal numbers: 00 binary 5 0 in decimal, 01 binary 5 1 in decimal, 10 binary 5 2 in decimal, and 11 binary 5 3 in decimal. A low logic state is 0 in binary, and a high logic state is 1 in binary. The binary count sequence as shown in Figure 12-21 is: 00, 10, 11, 01, 00

. . . Thus, there is a four-state repeating pattern, which translates into 0, 2, 3, 1, 0, . . . When we look at the output pins of U1 for C0, C1, C2, and C3, the numeral following the “Cx” is the state of the address line. So, the sequence of 0, 2, 3, 1 translates to a sequence of C0 first, C2 second, C3 third, and C1 fourth. Because 0 degrees  C0, 90 degrees  C2, 180 degrees  C3, and 270 degrees  C1, we see that the sequence is correct in having 0, 90, 180, and 270 degrees in order.

FIGURE 12-21 Top waveform at pin 14 (A input control line LSB) and bottom waveform at pin 2 (B input control line MSB) of the FST3253 analog switch with a 00, 01, 11, and 10 binary sequence or equivalently a 0, 2, 3, 1 order.

DC Bias Conditions

Confirm that pin 14 and pin 2 of U1 have the ~ 5-volt peak to peak waveforms as shown in Figure 12-21 when V_CLK is a 5-volt peak to peak square wave signal at

14.400 MHz.

Make sure the connections in U1 are connected as shown in the schematic in Figure 12-20, and that there is 15 volts at pin 16 and that pin 8 is tied to ground. Confirm that there is 2.5 volts DC at pins 7 and 9. If you do not have this, check the R1 and R2 voltage divider circuit and that there is 2.5 volts at C2 as shown in the schematic.

With Vin disconnected for now, measure the DC voltages at the hold capacitors C4, C5, C6, and C7 that should all measure about 2.5 volts within 15 percent. Also measure the DC output voltage at pin 6 of U3 and U4, which should be ~ 2.5 volts DC within 15 percent. Also confirm that the power supply decoupling capacitor, C3, has short leads (less than 0.5 inch) and is very close to pin 16 of U1. Likewise, decou- pling capacitor C13 should have short leads and be close to pin 14 of U2A.

Testing Circuit with an RF or Function Generator

Again, with the generator at V_CLK set to 14.4000 MHz, with R7 5 R8 5 47Ω and with a 240-mV peak to peak sine wave signal 3.600900 MHz set for 50Ω source resis-

tance for Vin, you should see 900 Hz IF signals at V_0 and V_90 that are 90 degrees apart.

Confirm large amplitude signals (e.g., 6.0 to 8.0 volts peak to peak) at Vout_I and Vout_Q that are the same amplitude and 90 degrees apart. See Figure 12-22 where the amplitudes are ~ 7.3 volts peak to peak.

FIGURE 12-22 V_I and V_Q output signals (~ 900 Hz) from top and bottom waveforms with ~ 7.3 volts to 7.5 volts peak to peak output, and with a 92-degree phase shift between the two waveforms.

However, does Figure 12-20’s circuit really work the way it should with two of its hold capacitors, C4 and C6, coupled to the low-resistance input terminals via R7 and R8? We should see signal from the hold capacitors C5, C7, C4, and C6 waveforms at 0, 90, 180, and 270 degrees. Just because we see the final result at V_I and V_Q that seems to look correct with about 90 degrees shifted apart in Figure 12-23, it does not mean everything is working correctly. We need to check further back and probe the hold capacitors. See Figure 12-23.

In Figure 12-23 the first and second waveforms should be 90 degrees apart, which is not the case. Also, the amplitudes of all waveforms should be equal, which is not the case.

Clearly, the low resistance loading of C4 and C6 in Figure 12-20 has an effect of “messing” up the amplitudes and phases of the waveforms.

The order of the waveforms can be viewed from the top as the first followed below by the second, followed again below by the third, and the fourth at the bottom. As a matter of fact, the first and third waveforms just like the second and fourth waveforms should be 180 degrees apart or out of phase, but in fact they are nearly

identical in phase.

FIGURE 12-23 Signals from top to bottom at C5, C7, C4, and C6 that we expect the waveforms to be at 0, 90, 180, and 270 degrees. However, this is clearly not the case. Amplitudes are measured at 200 mV/division for all waveforms.

If we disconnect or remove R7 from C4 and R8 from C6 in Figure 12-20 so that all hold capacitors are either loading into a high-resistance input of the op amps or left open circuit we see the hold capacitors’ waveforms in Figure 12-24, which looks like what we expect.

FIGURE 12-24 Waveforms from top to bottom of C5, C7, C4, and C6 that show the expected 0-, 90-, 180-, and 270-degree phase shifts, and note all waveforms have the same amplitude.

From Figure 12-24, as long as we keep all hold capacitors, C5, C7, C4, and C6, loading into a very high resistance (e.g., > 1MΩ, or infinite resistance), we will get the correct results. The waveforms shown in Figure 12-24 were done by disconnect- ing R7 and R8 from capacitors C4 and C6. Note now that the third waveform is 180 degrees phase shifted from the first waveform, and that the fourth waveform is an

inverted version (180 degrees) of the second waveform. These inversions are what we expect as shown in the original concept via Figures 12-18 and 12-19.

Improving the “Original Design”

One way to improve the design is to use instrumentation amplifiers. This will pro- vide high-resistance inputs to all the hold capacitors. Note that the 0.27-μf hold capacitors C4, C5, C6, and C7 will still yield only a < 10 kHz bandwidth, but they can be reduced in capacitance to as low as 1000 pf, which will provide close to 100 kHz IF bandwidth. See Figure 12-25 for one implementation that provides a high resistance load to all the hold capacitors.

Because we are using instrumentation amplifiers for U3 and U4, the DC voltages at Vout_I and Vout_Q will be close to 0 volts. This is because the DC voltages are about the same at all the hold capacitors C4, C5, C6, and C7. The instrumentation amplifiers will provide a pure subtraction of the DC voltages between V_0 and V_180 and also between V_90 and V_270, which leaves 0 volts at their output termi- nals (pins 8 and 9 of U3 and U4).

By replacing the LT1115 single op amps in Figure 12-20 with INA163 instru- mentation op amps in Figure 12-25, this circuit now works more ideally to what a sample-and-hold RF mixer is capable of. The IF bandwidth at V_0, V_90, V_180, and V_270 are now equal and can be controlled further by changing the capaci- tances of the hold capacitors, C4, C5, C6, and C7. For example, C4, C5, C6, and C7 can be changed from 0.27 μf to 1000 pf to provide a wider IF bandwidth. We can now compare the phases with the four signals at V_0, V_90, V_180, and V_270. Set the frequency of V_CLK to 14.4000 MHz at 5 volts peak to peak. Set an RF signal generator or function generator to 100 mV peak to peak with a source resis- tance of 50Ω at 3.600100 MHz (for an IF 5 100 Hz). Note that the unloaded signal from the generator will be twice the amplitude, or 200 mV peak to peak when observed connected directly to an oscilloscope oscilloscope with a 1MΩ or 10MΩ input resistance.

Observe with an oscilloscope the 100 Hz IF signals at the hold capacitors, C4, C5, C6, and C7 as shown in Figure 12-25. The voltages, V_0, V_90, V_180, and V_270 should be about 90 percent of the open circuit voltage of the generator. This will then be 90% 3 200 mV peak to peak or about 180 mV peak to peak at the hold capacitors.

Note because the IF bandwidth will be low, < 10 kHz, you can use a VFO (vari- able frequency oscillator, crystal controlled via a synthesizer chip) to tune across the ham radio band. Another way is to use a fixed oscillator for a particular band and tune just portion of it, such as a 96 kHz span in the 80-meter band. If you have a 192 kHz analog to digital converter sound card (which is somewhat expensive), you can

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FIGURE 12-25 An improved version with instrumentation amplifiers INA163 (U3 and U4) to provide equal high resistance loading to the hold capacitors C4, C5, C6, and C7.

tune up to 192 kHz across a ham radio band when C4, C5, C6, and C7 are changed to a lower capacitance such as 1000 pf.

To measure the output signals from the instrumentation amplifiers, reduce the amplitude of the signal generator from 100 mV peak to peak at 50Ω source resis- tance to 10 mV peak to peak at 50Ω source resistance and at 3.600100 MHz. The open circuit voltage will then be 20 mV peak to peak. Each capacitor voltage will have 90 percent of 20 mV peak to peak or 18 mV peak to peak. However, since the phases of the V_180 and V_270 are inverted phases of the signals from V_0 and V_90, the difference or subtraction of V_0 and V_180 will be twice the signal.

That is V_180 5 –V_0, so (V_0 – V_180) 5 (V_0 – –V_0) 5 (V_0 1 V_0) or (V_0 – V_180) 5 2 3 V_0.

With V_0 having 18 mV peak to peak, (V_0 – V_180) 5 2 3 V_0 5 2 3 18 mV peak to peak or (V_0 – V_180) 36 mV peak to peak.

Note that V_270 5 –V_90, and with V_90 5 18 mV peak to peak we have:

(V_90 – V_270) 5 (V_90 – –V_90) 5 (V_90 1 V_90) 5 2 3 V_90


(V_90 – V_270) 5 2 3 18 mV peak to peak or

(V_90 – V_270) 5 36 mV peak to peak

The outputs of the instrumentation amplifiers are approximately: (1 1 6000Ω/R7)(V_0 – V_180) and (1 1 6000Ω/R8)( V_90 – V_270)

In this case, R7 5 R8 5 62Ω and 6000Ω/62Ω 5 6000/62, so the outputs are: (1 1 6000/62)(V_0 – V_180) 5 97.7(V_0 – V_180), and 97.7(V_90 – V_270)

The expected outputs from the Vout_I 5 97.7(V_0 – V_180); Vout_I 5 97.7 (36 mV peak to peak) or Vout_I 5 3.5 volts peak to peak. Similarly, Vout_Q 5 3.5 volts peak to peak.

Because low noise instrumentation amplifiers are not always available, we can slightly modify Figure 12-20’s circuit at least two ways. See Figure 12-26 where the 180- and 270-degree signals are not used so that the op amps use only their high- resistance non-inverting input terminals for amplifying the 0- and 90-degree signals. In Figure 12-26 sample-and-hold capacitors C4 and C6 are not used and can be removed. However, all sample-and-hold capacitors are reduced in capacitance to 1000 pf to provide a wider IF bandwidth out to about 100 kHz.

For Figure 12-26’s circuit, the DC bias conditions should have 2.5 volts at pins 7 and 9 of U1 and also 2.5 volts DC at hold capacitors C5 and C7. Amplifiers U3 and U4 are configured as unity gain voltage followers for DC signals (e.g., for DC analy- sis, imagine removing the 100 uf capacitors C15 and C16). This means that the DC voltages at pin 6 of U3 and U4 will also be about 2.5 volts since the hold capacitor voltages at C5 and C7 are at 2.5 volts.

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FIGURE 12-26 A circuit requiring only the 0- and 90-degree signals via V_0 and V_90.

The AC gains for U3 and U4 are (1 1 R3/R7) or (1 1 R4/R8), which in this case is (1 1 4700/47), which leads to (1 1 100) 5 101.

With the same test conditions for Figure 12-25’s circuit where V_CLK is at

14.400 MHz and 5 volts peak to peak square wave, the RF or function generator provides Vin 5 10 mV peak for 50Ω source resistance with 20 mV peak to peak into a “no load” condition with frequency at 3.600100 MHz. The output amplitudes from Vout_I and Vout_Q should be 20 mV 3 90% 3 101 or 1.8 volts peak to peak. There should be a 90-degree phase shift on the Vout_Q referenced to Vout_I.

If we utilize all the signal voltages from the hold capacitors as shown in Figure 12-27, then we can about double the gain by adding voltage follower amplifiers U3A and U3B. These two voltage followers provide an equivalent instrumentation ampli- fier circuit. Via R7 and R8, the added voltage followers drive the inverting input terminals of U4A and U4B.

Not shown in Figure 12-27 are the connections to pins 7, 9, 14, and 2 of U1, which are connected to the circuits in Figure 12-26. For example, U1 pins 14 and 2 are connected to pins 6 and 9 of the U2A/B circuit in Figure 12-26. Also pins 7 and 9 of U1 in Figure 12-27 will be connected to an input circuit L1 and C1 as shown in Figure 12-26.

In this configuration, dual op amps are used. U3A and U3B are unity gain volt- age followers with U4A and U4B as voltage gain amplifiers. The voltage gains for the various AC signals across the hold capacitors are as follows:

Vout_I 5 (1 1 R3/R7) 3 V_0 1 –(R3/R7) 3 V_180

However, V_180 5 –V_0. This leads to:

Vout_I 5 (1 1 R3/R7) 3 V_0 1 –(R3/R7) 3 –V_0 which leads to:

(1 1 R3/R7) 3 V_0 1 (R3/R7) 3 V_0, or Vout_I 5 [1 1 2(R3/R7)] 3 V_0

With R3 5 7500Ω and R7 5 150Ω, Vout_I 5 [1 1 2(7500/150)] 3 V_0 or

Vout_I 5 (1 1 100) 3 V_0

Vout_I 101 V_0

Vout_Q 5 (1 1 R4/R8) 3 V_90 1 –(R4/R8) 3 V_2700. But V_270 5 –V_90.

Likewise, from the same type of calculations for Vout_I:

Vout_Q 5 [1 1 2(R4/R8)] 3 V_90

Vout_Q 5 101 3 V_90

With a Vin at 10 mV peak to peak signal for 50Ω source resistance (20 mV peak to peak into an open circuit and at 3.600100 MHz with V_CLK being a 14.4000 MHz 5-volt peak to peak square wave, the amplitude of Vout_I 5 20 mV 3 90% 3 101 5

1.8 volts peak to peak.

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FIGURE 12-27 Partial schematic with LM833 dual op amps that have 15-MHz gain bandwidth products, which replace instrumentation amplifiers (e.g., INA163 in Figure 12-25).

Similarly, the amplitude of Vout_Q is also 1.8 volts peak to peak and phase shifted 90 degrees referenced to Vout_I with sine waves at 100 Hz.

In terms of the DC voltages at pins 1 and 7 of U4A/B, there are 2.5 volts DC at all the hold capacitors C4, C5, C6, and C7 so we have the following:

U4A pin 1 DC 5 (1 1 R3/R7) 3 2.5 v 1 –(R3/R7) 3 2.5 v, which results in: U4A

pin 1 DC 5 2.5 v 1 (R3/R7) 3 2.5 v 1 –(R3/R7) 3 2.5 v U4A pin 1 DC 5 2.5 volts.

This is because the (R3/R7) 3 2.5 v terms cancel out.

Thus, U4A pin 1 DC 5 2.5 volts DC.

And similarly, U4B pin 7 DC 5 2.5 volts DC.

Another View of Op Amp Circuits

(Where the Inverting Input Drives a Load)

If we examine Figure 12-20’s circuit with C4 and R7 and also look at the waveforms of Figure 12-23, then we would expect the first and third waveforms to be 180 degrees apart, which they are not. Then the question is why? It turns out that the (1) input to a negative feedback amplifier actually provides a low-impedance “output” impedance to the (–) input terminal. The reason is that as long as the output signal of the negative feedback amplifier is not clipping to the power supply rail, then the (–) input’s voltage has to be about the same voltage as the (1) input terminal. The (–) input voltage follows the (1) input’s signal. See Figure 12-28.

FIGURE 12-28 Examining why the voltages V_0 and V_180 are not as expected as shown in this circuit, which is similar to part of Figure 12-20.

In a non-inverting gain op amp configuration, the input voltage, V_0, is applied to the (1) input terminal. Once this happens, the (–) input terminal must be “equal”

Some Ham Radio Circuits Related to SDR        303

to the (1) input. The output voltage at pin 6 U3 adjusts accordingly so that the volt- age at the (–) terminal matches the voltage of the (1) terminal. In essence, the signal to the (1) input terminal is the determining factor for “everything.” If you try to apply another voltage (e.g., V_180) to the (–) input via R7 or a series resistor, you will change the output voltage at pin 6, but you will not be able to the change the (–) input’s signal at pin 2. Thus at (–) input pin 2 we have V_0 due to the negative feed- back system via feedback resistor R3. Since pin 2 represents a pure voltage source of V_0 also, it provides a voltage to the right side of R7. So, going “backwards,” there is a V_0 voltage via R7 that travels into hold capacitor C4. The signal at C4 via V_0 overpowers the V_180 signal from the analog switch, so that C4 has some version of V_0 instead of V_180, where V_180 is a 180 degrees out-of-phase signal. So, the 180 degrees signal at C4 is “washed” out by the 0 degrees signal at C5. And the C4 signal looks similar in phase to the signal at C5. In fact, this is what we see when we com- pare the top first signal for C5 to the third signal for C4 in Figure 12-23.

FIGURE 12- 29 Redrawn Figure 12-28 for clarity as negative feedback amplifier with a series resistance R3 within the negative feedback loop.

As shown in Figure 12-29, you can make a voltage follower with a series output resistor such as R3 (e.g., 2200Ω) and still provide a low output resistance source of V_0 into the load, R7 and C4. The reason why ~ V_0 has a low resistance in spite of R3 being in series is that the negative feedback terminal (–) input is taken at ~ V_0. The negative feedback system then turns up Vout_I to compensate for loss due to the loading effects of R7 and C4 on R3. The true output resistance at the ~V_0 is approx- imately R3/a0(f), where a0(f) is the frequency-dependent gain of the open loop gain from U3. For example, if the LT1115 has a 50-MHz gain bandwidth product (GBWP), and we want to know the gain at f 5 1 kHz, then a0(f) 5 a0(1kHz) 5 GBWP/f 5 50 MHz/1kHz or 50,000. Then output resistance at ~V_0 is about R3/50,000 or 2200Ω/50,000 5 0.044Ω < 1Ω. We can see now that ~V_0 has low-output resistance and for all practical purposes, ~V_0 is a voltage drive signal into R7 (“Load”) as shown in Figure 12-29. So that is why in Figure 12-23, C4 has a voltage that looks more like V_0 instead of C4 having a signal that is out of phase (e.g., V_180).

Suggested System Approach

In Figures 12-20, 12-25, 12-26, and 12-27, the input signal Vin (e.g., an antenna sig- nal) goes directly to the mixer chip such as an FST3253. A better way is to take the matching network from Figure 12-7 with Cvar1, L1, R13 (e.g., R13 5 5100Ω for a Q

~ 10), R12, Q1, R10, R11, and C9. C9 from Figure 12-7 then feeds the Vin of Figures 12-20, 12-25, 12-26, and 12-27. This matching network steps up the antenna signal like a step-up transformer so that the antenna’s signal level is sufficiently large such that medium low noise op amps such as NE5523 or LM833 can be used. Secondly, the RF matching network via Cvar1 and L1 forms a filter network that reduces out- of-band signals and noise that would be mixed down to the IF signal.

Crystal Oscillators

For many amateur radio circuits a very stable frequency oscillator usually points toward a crystal resonator, although sometimes a ceramic resonator can be used. The order with the best to worst frequency stability for oscillator circuits we have: crystal, ceramic, inductor-capacitor (LC), and lastly resistor–capacitor (RC).

Before we look into some oscillator circuits, let’s take a look at a crystal or ceramic resonator. See Figure 12-30 of different types of crystal and ceramic resonators.

FIGURE 12-30 Cylindrical crystals or tuning forks on the left side, standard size crystals HC-49/S (shorter version) and HC-49/U (standard “longer” size) in the center, and a ceramic resonator on the right side that includes a three-lead ceramic resonator with built-in 15 pf capacitors.

The models of crystal or ceramic resonators are shown in Figure 12-31 with some circuits.

As shown in Figure 12-31(a), a crystal can be modeled as a series resonant circuit with inductor Ls, capacitor Cs, and series resistor Rs. Many oscillators are config- ured to use a crystal as a series resonant circuit, as shown in Figures 12-31(b), (c), and (d). Figure 12-31(a) shows the lead body capacitance, Co, which then also forms a parallel resonant circuit with Ls, Cs, and Rs. However, as we will see, most crystal oscillators run in some form of a series resonant mode.

Tekstvak: 305

FIGURES 12-31(A) TO (D) Schematic representation of a crystal or ceramic resonator Y1 in (a), a two-gate crystal/resonator oscillator in (b), a single-gate crystal/resonator oscillator in (c), and an oscillator where a three-terminal resonator Y1 includes internal capacitors C2 and C3 in (d).

Let’s first take a look at Figure 12-31(b), a two-gate oscillator circuit. Because there are two inverters (U1B and U1C) in series, the net phase shift is 180 degrees plus 180 degrees, or 360 degrees, which is like 0 degrees. Oscillation then occurs when the crystal acts like a band-pass filter with zero degrees phase shift at the reso- nant or crystal frequency. With this circuit, generally, capacitor C1 has a large value such as 0.01 μf. However, C1 may be selected to a different value to adjust the oscil- lation frequency. C1 should be larger than 15 pf or a parasitic oscillation may occur. In Figure 12-31(b), the first inverting gate, U1B has a feedback resistor, R1, that forms a low input resistance at pin 3. The R1 resistance value is chosen to be in the few hundreds to few thousands of ohms depending on the oscillation frequency. U1B via R1 has been configured as an analog inverting gain amplifier with low- resistance input. If the crystal, Y1, is removed, the output of the first inverter with R1 will sit at one-half the supply voltage (e.g., 2.5 volts) for 74AC (up to ~ 30 MHz), HC (up to ~ 10 MHz), and 4000 series CMOS chips (less than 1 MHz). When the crystal Y1 is put back in, it completes a positive feedback loop with the maximum gain at the crystal’s resonant frequency determined by Ls and Cs in Figure 12-31(a) that form a series resonant circuit within the crystal. This series resonant circuit in Y1 provides zero phase shift and maximum gain (amplitude) at resonance. For series mode crystals, the resonant frequency is just the one imprinted on the package. And for crystals with a load capacitance, the series capacitor C1 is set to the load capaci- tance (e.g., 18 pf) to tune the crystal to the specified frequency. C1 allows “tweaking” the oscillation frequency further even for series mode crystals, which means C1 can have a range of 0.1 uf to about 100 pf. Note that R2 and C2 along with C3 are included

in Figure 12-31(b) to prevent parasitic high frequency oscillations.

If we now look at Figure 12-31(c), we see a single inverter gate oscillator config- ured as an analog inverting amplifier. R1 is generally a very high resistance value ≥ 1MΩ to avoid loading down the oscillation signal at C3. The crystal Y1 is modeled as a series resistor inductor capacitor circuit with C3, the load capacitance of the crystal (e.g., 11 pf to 33 pf) that forms a low-pass filter. This low-pass filter at C3 provides 90 degrees to 135 degrees of phase shift at the resonant frequency. To achieve oscillation, we need 180 degrees of phase shift from the output pin 2 to the input pin 1 of U2A. To add more phase shift, R2 and C2 form an RC low-pass filter. Generally, we set the

–3 dB cut-off frequency of the R2-C2 low-pass filter at about one-half the crystal frequency to provide about 60 degrees of phase shift, which will be sufficient for the circuit to oscillate. Typically, R2 is in the range of 470Ω and 2200Ω and C2 has a capacitance value equal to or larger than C3, the crystal’s specified load capacitance. In certain cases, R2 may be increased to about 5kΩ and R1  4.7MΩ to 10MΩ for low-frequency (cylindrical) crystals whose frequencies ≤ 200 kHz.

Figure 12-31(d) shows essentially the same circuit as Figure 12-31(c) with a three-terminal ceramic resonator having internal C2 and C3 capacitors. This type of ceramic resonator has three leads including center ground lead. This device is sym-

metrical, and the two outer leads can be reversed. The center lead stays connected to ground. Typical values for R2 range from 220Ω to about 3300Ω.

NOTE: All crystals are symmetrical or non-polarized, which means that they will perform the same function with their leads reversed.

Types of Crystals

When you purchase crystals, they are separated into two groups. The most common group is crystals that are specified with loading capacitances such as 11 pf to 33 pf. Common loading capacitances are in the 18 pf to 22 pf range. The oscillator shown in Figure 12-31(b) is a series resonant circuit; the loading capacitance is C1. In series with crystal Y1 is C1 to tune the crystal to the specified frequency. For example, in Figure 12-31(b) having U1B and U1C alternatively as 74AHC04, R1 5 3300Ω, R2 5 56Ω, C2 5 10 pf, C3 5 10 pf, and with Y1 5 3.0000 MHz specified with 30 pf loading so C1 5 30pf, the oscillation frequency at pin 6, Vout measured at 2.999971 MHz or about 29 Hz off, which is within the 10-ppm tolerance of the crystal. For a 3-MHz crystal 10 ppm (10 parts per million) yields a tolerance of ±(3 MHz 3 10/million), which leads to ±30 Hz.

Other oscillators that make use of the loading capacitance specification are shown in Figures 12-31(c) and 12-31(d). In these two figures, C3 is the loading capacitance value. So, if you have a crystal specified at 18 pf, then C3 5 18 pf. The other capacitor, C2, works with resistor R2 to form a low-pass filter that provides a lagging phase shift of at least 45 degrees but less than 90 degrees. Typically, the phase shift is set to about 60 degrees by having the cut-off frequency, fc 5 1/(2πR2C2), set to one-half the crystal frequency. For example, suppose you have a 7 MHz crystal, you can set fc 5 3.5 MHz. Now we have to find a C2, but what’s a good value for R2? R2 generally can be in the range of 470Ω to 4700Ω. So, we can just pick R2 5 1000Ω. Then this leads to C2 5 1/(2πR2fc) 5 1/(2π1000Ω 3 3.5MHz) 5 45.5 pf, or C2 ~ 47 pf. Actually, C2 in the range of 47 pf to 68 pf will work fine. In other cases, you can use identical capacitance values for C2 and C3 that equal the specified loading capacitance such as 18 pf or 22 pf while having R2 5 470Ω to 4700Ω.

What if you want more phase shift closer to 90 degrees by lowering the cut-off

frequency further to, for example, one-twentieth of the crystal frequency? This will make C2 about 10 3 47 pf or C2  470 pf. The oscillator may not oscillate because although the phase shift will be about 87 degrees, the amplitude is attenuated by twentyfold, so only about 5 percent of the signal will enter the crystal Y1 on the right terminal side. The inverter gate U2A configured as an inverting gain amplifier will have to have sufficient gain to make up the attenuation to ensure oscillation. A twentyfold attenuation may have too much loss for the circuit to oscillate.

The second group of crystals is the series resonant type, and there is no loading capacitance specification. Basically, it means if you were to put a series mode crystal, Y1, in Figure 12-31(b) with C1 shorted out or C1 is from 0.01 μf to 0.1 μf (e.g., an AC short-circuit at the oscillation frequency), the oscillator will oscillate at the specified frequency.

Low-Frequency Cylindrical Crystals, “Standard” Crystals, and Ceramic Resonators

In the twenty-first century, most low-frequency crystals at ≤ 200 kHz are the types shown in Figure 12-30 (left side). These provide accurate frequency references for clocks, watches, and other devices. A very popular frequency 32.768 kHz is used in almost all watches. These low frequency crystals may have about a thousand times more loss due to Rs ~ 50kΩ as shown in Figure 12-31(a), whereas the larger standard size HC-49/S and HC-49/U crystals have Rs ~ 50Ω or less. Typically, the load capac- itance of the low-frequency smaller cylindrical crystals is in the 11 pf to 22 pf range. Also, they must be driven with lower AC currents when compared to the standard crystals or ceramic resonators. So, you can more easily damage a low-frequency cylindrical crystal if you are not careful.

A low-frequency oscillator such as in Figure 12-31(c) has the low-pass filter section R2 5 4700Ω, C2 5 1000 pf, Y1 5 32.768 kHz to 100 kHz, C3 5 15 pf, and R1 5 4.7MΩ. It is important to raise R1 to 4.7MΩ so that the lossy Rs 5 50kΩ inside the cylindrical crystal transfers enough AC signal into the input terminal of the inverter gate, U2A that has been converted to an inverting gain amplifier. Note: If R11MΩ, this resistance value may be too low; and the circuit might not oscillate with cylindrical crystals whose frequencies are ≤ 200 kHz. Given the low oscillation fre- quency, inverter U2A may be a 74C04 or CD4069 for slower rise/fall times that do not propagate output signal glitches, or you can still use the faster 74HC04 chip.

Standard HC-49 and High-Frequency Cylindrical Crystals

Standard crystals such as HC-49/S and HC-49/U have series resistance, Rs < 100Ω and with that generally as shown in Figure 12-31(c), R2 can be in the 470Ω to 2200Ω range. Again, R1 can be in the 1MΩ to 4.7MΩ range. Typically, R1 5 2.2MΩ. In general, C2 should be at least the loading capacitance, if not more. For example, sup- pose the crystal frequency is 20 MHz with an 18 pf load capacitance. We should choose a fast inverter gate beyond a 74HC04 such as a 74AC04, 74AHC04, or 74VHC04. C3 will equal the load capacitance of 18 pf, and if R2 5 470Ω, then C2 5 33 pf for an R2/C2 low-pass filter cut-off frequency, fc 5 1/(2πR2C2). We set fc 5 [0.5 3 crystal frequency], and this will be 10MHz, half of the crystal’s 20-MHz frequency to provide about 60 degrees of phase shift at 20 MHz.

Note there are fundamental frequency crystals above 200 kHz such as ≥ 3MHz that are available in cylindrical packages like the 32.768 kHz types. However, these cylindrical crystals have the series resistance, Rs, much lower at ≤ 200Ω, which allows them to be used in crystal oscillator circuits with essentially the same R2, C2, and C3 values like the HC-49/S and HC-49/U crystals.

In terms of frequency accuracy, crystals are rated in parts per million (ppm) and can range from 10 ppm to 100 ppm. For good accuracy, you can specify for better (less) than 30 ppm. But just remember the more frequency accuracy you require, the more it may cost. Fortunately, in some cases, you can replace C3 in Figure 12-31(c) with a variable trimmer capacitor (e.g., 5 pf to 35 pf) to adjust the frequency.

Ceramic Resonators

There are generally two types of ceramic resonators, low frequency and higher fre- quency types. The lower frequency ones resonate below 1.3 MHz with load capaci- tances in the 100 pf to 560pf. For the resonators above 1.3 MHz, the load capaci- tances are from 15 pf to 33 pf, which is similar to crystals. Their equivalent series resistance is < 500Ω, so they can be used very much like standard crystals. Ceramic resonators are generally used in situations when the required frequency accuracy is better than an LC or RC oscillator, but does not need to be as accurate as a crystal. Ceramic resonators find their way in almost all remote controls. The frequency tol- erance rating is in the order of 0.3 percent to 0.5 percent or 3000 ppm to 5000 ppm. There are some ceramic resonators that have accuracies in the order of 0.1 percent or 1000 ppm, such as the surface mount package Murata CSTNR_GH5L (4.00 MHz to 7.99 MHz), CSTNE_GH5L (8.00MHz to 13.99 MHz), and CSTNE_VH3L (14.00 MHz to 20.00 MHz).

As an example, using instead a 74AHC04 inverter gate in Figure 12-31(c), set R154.7MΩ,R252200Ω,andwithY15455kHz(e.g.,MurataCDBLA455KCAY16-B0),

set the cut-off frequency fc 5 1/(2πR2C2) 5 0.5 3 455 kHz 5 227.5 kHz. This leads to:

C2 5 1/[2π2200Ω(227.5 kHz)] or C2 5 318 pf ~ 330 pf

Initially set C3 to 330 pf, then try different values from 100 pf to 560 pf to trim the oscillation frequency to 455 kHz using a frequency counter set to the 1MΩ input resistance mode. In this 1MΩ input resistance mode, you can generally use a 310 scope probe to measure the oscillation frequency if there is sufficient output signal (e.g., > 2 volts peak to peak).

NOTE: Some frequency counters have a 50Ω input resistance mode, which will load down the oscillator’s signal.

If C3 is replaced with a variable capacitor, such as 10 pf to 365 pf air dielectric type or a poly varicon (e.g., a 10 pf to 140 pf or a 10 pf to 270 pf), then this 455-kHz oscillator may be used for product detectors or beat frequency oscillators with vari- able pitch control.

Finally, Figure 12-31(d) shows a ceramic resonator circuit with built-in capaci- tors C2 and C3. For some ceramic resonators, the capacitances are 15 pf each. Other ceramic resonators may be up to 47 pf for their internal C2 and C3. As always, you can add parallel capacitor(s) across the internal C2 and/or C3 to adjust the fre- quency. Or a variable trimmer capacitor such as 5 pf to 35 pf can be placed between the inverter’s input pin and ground. The feedback resistor R1 may be in the 1MΩ to 4.7MΩ range. Again, set the R2 resistance to provide a phase shift with the internal C2. For example, if the ceramic resonator is at 7.2 MHz and the internal C2 and C3 are 15 pf each, R2 can be set as: R2 5 1/[2π15pf(7.2MHz/2)] or R2 5 1474Ω ~ 1500Ω. Again, the cut-off frequency is set at one-half the oscillation frequency of 7.2 MHz, which results in 3.6 MHz (7.2 MHz /2).

We now will look at a one-transistor oscillator with a logic level shifting ampli- fier that is commonly used when a 0- to 5-volt logic signal is required to drive logic circuits including flip flops, gates, etc. See Figure 12-32.

FIGURE 12-32 A “typical” one-transistor emitter follower crystal oscillator Q1 with common emitter amplifier Q2 to provide logic level signals for 5 volts peak to peak output.

C5 is generally the crystal’s specified load capacitance such as 18 pf. However, C5 may be replaced with a 5 pf to 35 pf trimmer variable capacitor to adjust the fre- quency exactly at Vout1 using a frequency counter. If Y1 is a series resonant crystal, C5 5 0.01 uf, which is close to an AC short-circuit at the crystal’s frequency.

So how does the Q1 crystal oscillator work? Q1 is an emitter follower, which means it has a voltage gain of about 1. Yet somehow this circuit must step up the signal voltage at the resonant frequency of the crystal, Y1. At the emitter, you will notice a small value capacitor C1 that forms a very slight low-pass filter effect. That is the signal at Q1’s emitter has a high-frequency roll-off compared to Q1’s base sig- nal. The high-frequency roll-off at the emitter also causes a phase lag with respect to the signal at Q1’s base. In order to get an oscillation going, we need C3 and Y1 to form a high Q high-pass filter that has a peaked amplitude response near the crys- tal’s frequency. The signal voltage at VY1 or base of Q1 is a signal that is stepped up and has a phase lead referenced to the signal at the emitter of Q1. Crystal Y1 forms an inductive impedance at near the crystal frequency. By near the crystal frequency, we mean within less than 0.1 percent tolerance of the crystal frequency, typically within 0.01 percent or better. For example, if the crystal frequency is 10.000 MHz, the oscillator will provide signal at 10.000 MHz within 1.0 kHz. For example, 1 kHz is 0.01 percent of 10.000 MHz or 100 ppm of 10.000 MHz where ppm 5 parts per million, and 100/million 3 10 million Hz 5 1 kHz. Because there is a phase lead due to the crystal and a phase lag via C2, the net phase shift is 0 degrees. And the stepped- up signal via the high Q high-pass filter sustains an oscillation. The actual signal at VY1 (or VE1) is generally not always a sine wave, but instead it can be a distorted waveform. See Figure 12-33.

To troubleshoot this circuit, we first look at the DC conditions, which are easier

to measure when the crystal Y1 is removed. The reason for this is when a circuit generates high-frequency AC signals, it causes measurement problems when using a DVM.

A first approximation for Q1’s base voltage is determined by the resistive voltage divider circuit R2 and R3. Since they are equal, the Q1 base DC voltage should be about half of the 5-volt supply voltage, or 2.5 volts DC. The emitter voltage then should be about 0.7 volt below the base voltage. So, the DC emitter voltage of Q1 should be 2.5 volts – 0.7 volt or about 1.8 volts.

In general, there will be some base current from the transistors due to the cur- rent gain, β, not being “infinite.” So, the emitter voltage can be as low as 80 percent of the calculated value (e.g., if the emitter voltage is calculated to be 1.8 volts, the base currents may cause the emitter voltage to drop to 1.8 volts 3 0.80 or 1.44 volts). For example, Q1 5 2N3904 and Q2 5 2N3906, where β for each transistor is generally ≥ 100. If by chance Q1’s base voltage is much lower than 2.5 volts, such as 1 volt or less, check to see if the transistor’s collector and emitter leads are installed correctly. If the collector and emitter terminals are reversed, the current gain, β, drops to < 5 and there will be excessive base currents drawn through R2 and R3 that

will produce a much lower DC voltage at Q1’s base terminal.

There is a base series resistor, R1 at 47Ω, to prevent parasitic high-frequency oscillations. Without it, or R1 5 0Ω, there can be a “risk” of the oscillator working incorrectly. Also note that L1 and C1 provide localized power supply decoupling so

that any power supply glitches or noise does not induce noise into the 15-volt sup- ply line that may be powering other circuits.

With the crystal Y1 reinstalled, you will need an oscilloscope next to trouble- shoot this circuit. The values for C2 and C3 depend on the crystal’s frequency. See Table 12-1 for suggested values for Figure 12-32.

NOTE: The crystal Y1 is a fundamental frequency crystal. Be sure to look out for that and not use an overtone frequency crystal.

TABLE 12-1  Suggested Capacitor Values Based on the Crystal’s Frequency

Crystal FrequencyC2C3C4R4
32 kHz to 200 kHzDo not useDo not useDo not useDo not Use
1 MHz to 5 MHZ470 pf470 pf100 pf470Ω
5 MHz to 15 MHz47 pf100 pf47 pf470Ω
15 MHz to 30 MHz22 pf100 pf47 pf470Ω
455 kHz Cer. Res.470 pf1000 pf100 pf470Ω

The signals from the oscillator section coupled to the base of the amplifier, Q2, and the final output signal at Q2’s collector are shown in Figure 12-33. Notice that the output signal is not necessarily always a nice pulse signal.

FIGURE 12-33 Top trace is a ~ 2-volt peak to peak 28 MHz signal at Q2’s base or VY1; the bottom waveform shows 5 volts peak to peak 28 MHz at Vout1 for Figure 12-32’s circuit.

An alternative circuit using DC feedback with a filter capacitor can also be implemented. The advantage of this circuit is that the oscillator and amplifier can work with a range of voltages without having to change value for Q2’s base biasing resistors. See Figure 12-34.

FIGURE 12-34 Automatically biasing Q2 for a range of supply voltages via resistors R5 and R6 with feedback filter capacitor C5; use the suggested C2, C3, C4, and R4 values in Table 12-1.

C6 is generally the crystal’s specified load capacitance such as 18 pf. However, C6 may be replaced with a 5 pf to 35 pf trimmer variable capacitor to adjust the fre- quency exactly at Vout1 using a frequency counter. If Y1 is a series resonant crystal, C6 5 0.01 uf, which is close to an AC short-circuit at the crystal’s frequency.

Q2’s base voltage is biased to the supply voltage –0.7 volt and in this example it is 5 v – 0.7 v or 4.3 volts at Q2’s base. For troubleshooting, we can safely probe the DC voltage at C5, which will have a DC voltage close to 4.3 volts, but likely to be in the range of 0.5 volt to 1.2 volts lower than Q2’s base voltage.

So C5’s DC voltage may be in the range of:

(4.3 volts – 0.5 volt) 5 3.8 volts to (4.3 volts – 1.2 volts) 5 3.1 volts

Again, you can have Q1 as 2N3904 and Q2 as 2N3906, or equivalent transistors. Generally, these transistors will work fine for oscillation frequencies up to 30 MHz and possibly more.

Of course, we can use a single logic gate inverter chip to make a crystal oscillator with logic-level output. We first encountered this in Figure 12-5, IC1E (74HC04 hex inverter gate) and in Figure 12-31(c). Let’s take another look at these types of oscil-

lators. In Figure 12-35(a) we see a very common single-gate inverter crystal oscilla- tor with a DC bias feedback resistor R1, which is usually between 1MΩ and 4.7MΩ. The reason for the high resistance value is to avoid loading down the signal at C5 via the crystal Y1.

NOTE: The crystal Y1 acts like an inductor and forms a two-pole, low-pass filter with C5 as a voltage divider with the internal equivalent series capacitor in Y1. This two- pole, low-pass filter provides another 90 degrees to 150 degrees of phase lag to combine with R8 and C4 so that there is a total of 180 degrees of phase shift that will enable oscillation. If you want to confirm that Y1 acts like an inductor to enable oscillation, you can substitute an inductor such as 1 μH to 10 μH in place of Y1 in Figure 12-31(c) and you will see that the circuit oscillates.

For example, if R1 has a smaller resistance value such as 470Ω, the oscillation signal at C5 will be attenuated to the point where no oscillation occurs. R1 is to bias the inverter gate to become an inverting gain amplifier with a “reasonable” high input resistance as seen by C5 looking into R1 and IC1A’s input pin 1. For 74ACxx, 74HCxx, 74AHCxx, 74VHCxx, and 4000 series CMOS gates, R1 biases the input pin 1 to one-half the supply voltage such as 2.5 volts DC to provide close to 50 percent duty cycle pulses at output pin 2.

The problem with Figure 12-35(a) is that C4 is used to provide a phase lag of at least 45 degrees (via a –3 dB cut-off frequency) with the internal output resistance of IC1A. However, with faster chips such as the 74HC or 74AC logic families, the inter- nal resistance is so low that it may not provide enough phase lag, and therefore no oscillation occurs. The reason why Figure 12-35(a) works is because the crystal fre- quency is high enough (e.g., > 14 MHz) such that the gate delay of the inverter IC1A plus the output resistance with C4 has enough phase lag to ensure oscillation. If you try a lower frequency crystal in the ≤1 MHz range, chances are that the reliability of oscillation will be spotty. To examine further the phase lag requirement and role from C4, please see Figure 12-35(b).

A model of the crystal oscillator is shown in Figure 12-35(b) where Rout is the internal output resistance of the logic inverter gate. In the first generation CD4000 series CMOS (Complementary Metal Oxide Silicon) logic chips, Rout was in the order of 500Ω to 2000Ω. Thus, the phase lag network was approximated by a Rout-C4 low-pass filter with a – 3dB cut-off frequency fc 5 1/[2π(Rout C4)]. If Rout 5 1800Ω and C4 5 22 pf, then fc 5 4 MHz for a 5-volt supply. However, if we used a faster logic gate, such as a 74AC04 that has Rout < 10Ω, having C4 at 22 pf will not provide enough phase lag. We would have to either increase C4 to something like 4700 pf that will then cause the output pin to waste power driving C4’s low impedance, or better yet we can reduce power by adding a series resistor R8 as shown in Figure 12-35(c). By having R8 in the 200Ω to 2700Ω range, we can ensure that C4 provides the necessary phase lag for oscillation. In general, the inverter’s input capacitor C5

Tekstvak: 315

FIGURE 12-35 (a) A simple and not so reliable version of the one-gate oscillator, (b) a model of the oscillator, and (c) a more reliable design with series resistor R8.

can be set to the crystal’s loading capacitance, which is specified generally between 11 pf and 39 pf, with common loading capacitances of 18 pf and 22 pf for C5. See Table 12-2 for suggested R8 and C4 values, types of inverter gates, and types of reso- nators Y1 that can include crystal or ceramic resonators.

TABLE 12-2 Suggested Values for Figure 12-35(c) Using Fundamental Frequency Crystals and Resonators Having Load Capacitance

Crystal FrequencyC4C5*R1R8Logic Family
32 kHz to 100 kHz1000 pf15 pf4.7MΩ4700ΩCD4069, 74C04, 74HC04
1 MHz to 2 MHZ150 pf18 pf2.2MΩ2200ΩCD4069, 74C04, 74HC04
2 MHz to 4 MHz150 pf18 pf2.2MΩ1000Ω74HC04
4 MHz to 8 MHz68 pf18 pf2.2MΩ1000Ω74HC04
8 MHz to 15 MHz68 pf18 pf2.2MΩ470Ω74HC04,74HCU04,74AHC04
15 MHz to 30 MHz47 pf18 pf2.2MΩ470Ω74AC04, 74AHC04,74VHC04
Ceramic Resonators
455 kHz to 503 kHz330 pf†100 pf†2.2MΩ2200Ω74HC04, 74AHC04
3.0 MHz to 6.0 MHz100